Answer:
Equation: [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]
Intersection: [tex](y-2)^2+(z-5)^2=7[/tex]
Step-by-step explanation:
We are asked to write an equation of the sphere with center center [tex](-3,2,5)[/tex] and radius 4.
We know that equation of a sphere with radius 'r' and center at [tex](h,k,l)[/tex] is in form:
[tex](x-h)^2+(y-k)^2+(z-l)^2=r^2[/tex]
Since center of the given sphere is at point [tex](-3,2,5)[/tex], so we will substitute [tex]h=-3[/tex], [tex]k=2[/tex], [tex]l=5[/tex] and [tex]r=4[/tex] in above equation as:
[tex](x-(-3))^2+(y-2)^2+(z-5)^2=4^2[/tex]
[tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex]
Therefore, our required equation would be [tex](x+3)^2+(y-2)^2+(z-5)^2=16[/tex].
To find the intersection of our sphere with the yz-plane, we will substitute [tex]x=0[/tex] in our equation as:
[tex](0+3)^2+(y-2)^2+(z-5)^2=16[/tex]
[tex]9+(y-2)^2+(z-5)^2=16[/tex]
[tex]9-9+(y-2)^2+(z-5)^2=16-9[/tex]
[tex](y-2)^2+(z-5)^2=7[/tex]
Therefore, the intersection of the given sphere with the yz-plane would be [tex](y-2)^2+(z-5)^2=7[/tex].
An urn contains 4 blue balls and 6 orange balls. In how many ways can we select 2 blue balls and 5 orange balls from the urn? a) 24 b) 38 c) 36 d) 732 e) 8637 f) None of the above.
Answer:
b) 36
Step-by-step explanation:
We can use combinations to solve this problem.
The binomial coefficient [tex]\binom{n}{k}=\frac{n!}{k!(n-k)!}[/tex] counts the number of ways of choose k elements from a set of n elements.
The product rule from combinatorics says that if there are N ways of doing something and M ways of doing another thing, the number of ways of doing both things is equal to NM.
First, we choose the blue balls. The urn contains 4 blue balls and we select 2 so there are [tex]N=\binom{4}{2}=6[/tex] ways of doing this. Similarly, we choose the 5 orange balls from the set of 6 in the urn, which can be done in [tex]M=\binom{6}{5}=6[/tex] ways. By the product rule, there are MN=6(6)=36 ways of selecting all the balls.
The number of ways 2 blue balls and 5 orange balls can be selected from the urn is 36 number of ways: Option C is correct
Combination has to do with selection.
If r number is selected from n number, this is expressed using the formula:
[tex]nC_r=\frac{n!}{(n-r)!r!}\\[/tex]
If 2 blue balls are selected from 4 blue balls, this is expressed as:
[tex]4C_2=\frac{4!}{(4-2)!2!}\\4C_2=\frac{4\times 3 \times 2!!}{2!2!}\\4C_2=\frac{12}{2} = 6 ways[/tex]
Similarly, if 5 orange balls are selected from 5 orange balls, this is expressed as:
[tex]6C_5=\frac{6!}{(6-5)!5!}\\6C_5=\frac{6\times 5!}{1!5!}\\6C_5=\frac{6}{1} = 6 ways[/tex]
The number of ways 2 blue balls and 5 orange balls can be selected from the urn is 36 number of ways
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A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)=72,000+60x and p(x)=300−(x/20),
0l≤x≤6000.
(A) Find the maximum revenue.
(B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set.
(C) If the government decides to tax the company $55 for each set it produces, how many sets should the company manufacture each month to maximize its profit? What is the maximum profit? What should the company charge for each set?
By deriving and equating to zero the revenue and profit functions, the maximum revenue is $900,000 achieved at 3000 units sold. The maximum profit is $384,000 when 2400 units are sold at $180 each. If taxed $55/unit, the maximum profit is $302,400 at 1840 units with a price of $208/unit.
Explanation:To solve this, we'll first calculate the revenue function R(x) which is the product of the number of units sold and the price per unit, i.e., R(x) = x*p(x). Then, we'll find the profit function P(x), which is the difference between the revenue and the cost, i.e., P(x) = R(x) - C(x). Next, to maximize revenue and profit, we'll get the derivative of both R(x) and P(x), set them equal to zero, and solve.
For part (A), R(x) = x*(300 - x/20) = 300x - x^2/20. Its derivative is R'(x) = 300 - x/10. Setting R'(x) = 0, we get x = 3000 units, which leads to the maximum revenue of $900,000.
For part (B), P(x) = R(x) - C(x) = 300x - x^2/20 - (72000 + 60x). Its derivative, P'(x), when equal to zero gives us x = 2400 units. Thus, the maximum profit is $384,000 and the price per unit is $180.
For part (C), the new cost function very becomes C(x) = 72,000 + 115x. Setting the derivative of the new profit equation P'(x) = 0, we get x = 1840 units, which leads to the maximum profit of $302,400. The company should charge $208 per unit.
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(A) Maximum revenue: $450,000
(B) Maximum profit: $216,000 at production level of 2400 sets, price per set: $180
(C) With a $55 tax per set, maximum profit is $84,000 at production level of 2400 sets, price per set remains $180.
(A) To find the maximum revenue, we first need to maximize the revenue function:
[tex]\[ R(x) = (300 - \frac{x}{20}) \times x \]\[ R(x) = 300x - \frac{1}{20}x^2 \][/tex]
Now, let's find the critical points by taking the derivative of R(X) and setting it equal to zero:
[tex]\[ \frac{dR}{dx} = 300 - \frac{1}{10}x \]\[ \frac{1}{10}x = 300 \]\[ x = 3000 \][/tex]
So, the maximum revenue occurs at \( x = 3000 \).
Plugging \( x = 3000 \) back into the revenue function:
[tex]\[ R(3000) = (300 - \frac{3000}{20}) \times 3000 \]\[ R(3000) = (300 - 150) \times 3000 \]\[ R(3000) = 150 \times 3000 = 450000 \][/tex]
The maximum revenue is $450,000.
(B) To find the maximum profit, we need to maximize the profit function P(x) :
P(x) = R(x) - C(x)
Given C(x) = 72,000 + 60x, we have:
[tex]\[ P(x) = (300 - \frac{x}{20}) \times x - (72,000 + 60x) \]\[ P(x) = (300x - \frac{1}{20}x^2) - (72,000 + 60x) \]\[ P(x) = 300x - \frac{1}{20}x^2 - 72,000 - 60x \]\[ P(x) = -\frac{1}{20}x^2 + 240x - 72,000 \][/tex]
Now, let's find the critical points by taking the derivative of P(x) and setting it equal to zero:
[tex]\[ \frac{dP}{dx} = -\frac{1}{10}x + 240 \][/tex]
Setting dP/dx = 0:
-1/10x + 240 = 0
1/10x = 240
x = 2400
Now, we need to check the endpoints of the interval [tex]\( 0 \leq x \leq 6000 \)[/tex] for potential maximum profit.
[tex]\[ P(0) = -\frac{1}{20}(0)^2 + 240(0) - 72,000 = -72,000 \]\[ P(6000) = -\frac{1}{20}(6000)^2 + 240(6000) - 72,000 = -132,000 \][/tex]
So, the maximum profit occurs at x = 2400.
Plugging x = 2400 back into the profit function:
P(2400) = [tex]-\frac{1}{20}(2400)^2[/tex] + 240(2400) - 72,000
P(2400) = -288,000 + 576,000 - 72,000
P(2400) = 216,000
The maximum profit is $216,000.
To find the price per set at the optimal production level, we use the price-demand equation:
p(2400) = 300 - 2400/20
p(2400) = 300 - 120
p(2400) = 180
So, the company should charge $180 per television set to realize the maximum profit.
(C) If the government decides to tax the company $55 for each set it produces, the cost function becomes:
C(x) = 72,000 + 60x + 55x
C(x) = 72,000 + 115x
To find the new maximum profit, we repeat the steps from part (B) with the updated cost function. We already know that the optimal production level is x = 2400, so we can directly plug this value into the new profit function.
P(x) = R(x) - C(x)
P(2400) = R(2400) - C(2400)
Plugging in x = 2400 into the revenue function:
R(2400) = (300 - 2400/20 * 2400
R(2400) = (300 - 120) * 2400
R(2400) = 180 * 2400
R(2400) = 432,000
Plugging in x = 2400 into the updated cost function:
C(2400) = 72,000 + 115 * 2400
C(2400) = 72,000 + 276,000
C(2400) = 348,000
Now, calculate the new profit:
P(2400) = R(2400) - C(2400)
P(2400) = 432,000 - 348,000
P(2400) = 84,000
So, the company should manufacture 2400 sets each month to maximize its profit. The maximum profit is $84,000.
To find the price per set at the optimal production level with the tax:
p(2400) = 300 - 2400/20
p(2400) = 300 - 120
p(2400) = 180
The company should still charge $180 per television set to realize the maximum profit, even with the tax.
Therefore,
(A) Maximum revenue: $450,000
(B) Maximum profit: $216,000 at production level of 2400 sets, price per set: $180
(C) With a $55 tax per set, maximum profit is $84,000 at production level of 2400 sets, price per set remains $180.
According to a random sample taken at 12 A.M., body temperatures of healthy adults have a bell-shaped distribution with a mean of 98.28degreesF and a standard deviation of 0.63degreesF. Using Chebyshev's theorem, what do we know about the percentage of healthy adults with body temperatures that are within 2 standard deviations of the mean? What are the minimum and maximum possible body temperatures that are within 2 standard deviations of the mean? At least nothing% of healthy adults have body temperatures within 2 standard deviations of 98.28degreesF.
Answer:
At least 75% of healthy adults have body temperatures within 2 standard deviations of 98.28degreesF.
The minimum possible body temperature that is within 2 standard deviation of the mean is 97.02F and the maximum possible body temperature that is within 2 standard deviations of the mean is 99.54F.
Step-by-step explanation:
Chebyshev's theorem states that, for a normally distributed(bell-shaped )variable:
75% of the measures are within 2 standard deviations of the mean
89% of the measures are within 3 standard deviations of the mean.
Using Chebyshev's theorem, what do we know about the percentage of healthy adults with body temperatures that are within 2 standard deviations of the mean?
At least 75% of healthy adults have body temperatures within 2 standard deviations of 98.28degreesF.
Range:
Mean: 98.28
Standard deviation: 0.63
Minimum = 98.28 - 2*0.63 = 97.02F
Maximum = 98.28 + 2*0.63 = 99.54F
The minimum possible body temperature that is within 2 standard deviation of the mean is 97.02F and the maximum possible body temperature that is within 2 standard deviations of the mean is 99.54F.
A monkey has 200 bananas. He wants to transport as many bananas as possible to a destination which is 100 yards away. The monkey cannot carry more than 100 bananas at a time and he must eat a banana every yard it travels (regardless of the direction he walks). What is the maximum number of bananas that can be transferred to the destination?
Answer: 33 bananas
Step-by-step explanation:
Since the monkey can only carry 100 bananas at a time.
And a total of 200 bananas is to be transferred through 100yards.
With the condition that he eats 1 banana per yard .
If he decides to carry 100 banana straight through the whole 100yards. He would be left with no banana at the end of the journey and will not have enough banana to be able to come back and carry the rest.
Therefore, the best way is to transfer all the banana to a destination where he will be left with 100bananas that he can then move at once to the final destination.
To move 200 bananas to a particular destination given that he can carry only 100 banana each.
He would need to travel twice that means ( three trips i.e leave 100 return with 0 leave 0)
For three trips through a particular length in yards to exhaust 100 yards. The length in yards is given as
100 = 3 × l
l = 100/3
l = 33 yards
Travelling 33 yards he would be left with 100 bananas that he can carry in just one trip through the rest of the journey which is 100 -33 yards = 67yards
Carrying 100 bananas through 67 yards in one trip
He will exhaust;
1 banana per yard × 67yards = 67 bananas
Therefore, he would be left with
100 - 67 bananas
= 33 bananas
The Best Company produces two commercial products : blenders and mixers. Both products require a two step production process involving delivery of parts (JIT process) and assembly. It takes 1 hour to deliver parts for each blender and 2 hours for each mixer. Final assembly of mixers and blenders require 3 and 2 hours, respectively. The production capability is such that only 24 hours of delivery time and 30 hours of assembly time are available. Each blender produced nets the firm $7 and each mixer $6.
a) How many of each should be produced to maximize profit? Partial production of each product is allowed.
b) What is the $ amount of this profit?
Answer:
Maximum profit is $87 when 3 blenders and 11 mixers are produced.
Step-by-step explanation:
let blender is represented by [tex]x_{1}[/tex] and and mixer by [tex]x_{2}[/tex].
total time to deliver parts = 24 hrs
total time to assemble = 30 hrs
time taken by each blender to deliver parts = 1 hr
time taken by each mixer to deliver parts = 2 hr
time taken by blenders in final assembling= 2 hr
time taken by mixers in final assembling = 3 hr
Each blender produced nets the firm= $7
Each mixer produced nets the firm= $6
Using this all data linear system of equation will be:
[tex]x_{1} + 2x_{2} =24 ----- (1)\\2x_{1} + 2x_{2} = 30 ----- (2)\\[/tex]
profit function:
[tex]z= 7x_{1} +6x_{2} --- (3)[/tex]
[tex]from (1)\\x_{1} = 0 \implies x_{2}= 12\\x_{2}= 0 \implies x_{1}= 24\\[/tex]
Coordinate points obtained from (1) are (0,12) and (24,0)
[tex]from (2)\\x_{1}=0 \implies x_{2}=10\\x_{2}=0 \implies x_{1}=15\\[/tex]
Coordinate points obtained from (2) are (0,10) and (15,0)
plotting these on graph
points lying in feasible region are:
A(0,0)
B(0,10)
C(3,11)
D(12,0)
substituting these points in (3) to find the maximum profit:
for A (0,0)
z = 0
for B (0,10)
z = 60
for C (3,11)
z = 87
for D (12,0)
z=84
So maximum profit is $87 when 3 blenders and 11 mixers are produced.
A development economist is interested in whether average years of schooling of girls and boys are the same in a certain developing country. A random sample of 250 girls yields a mean of 5.1 years, and a standard deviation of 1.2 years. An independent random sample of 280 boys yields a mean of 6.3 years, and a standard deviation of 1.7 years.Test the null hypothesis that mean years of schooling is the same in the populations of girls (X population) and boys (Y population), against the alternative hypothesis that the population means are different. Use a 5% level of significance.1. What is the value of the lower end of the confidence interval?a. – 1.4086
b. – 1.4485
c. – 1.3715
d. – 1.40442. What is the value of the upper end of the confidence interval?a. – 1.2085
b. – 0.9956
c. – 0.9515
d. – 0.99143. What kind of distribution ( central limit theorem or T) and why?
Answer:
So on this case the 95% confidence interval would be given by [tex]-1.4485 \leq \mu_G -\mu_B \leq -0.9515[/tex]
Since the confidence interval not contains the 0 we can say that we have significant differences between the mean of girls and boys.
1. What is the value of the lower end of the confidence interval?
b. – 1.4485
2. What is the value of the upper end of the confidence interval?
c. – 0.9515
What kind of distribution ( central limit theorem or T) and why?
We can use the t distribution but since the sample size is large enough we will have a distribution similar to the normal standard distribution. Because when the degrees of freedom of the t distribution increases we have a normal distribution.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X_1 =5.1[/tex] represent the sample mean 1 (girls)
[tex]\bar X_2 =6.3[/tex] represent the sample mean 2 (boys)
n1=250 represent the sample 1 size
n2=280 represent the sample 2 size
[tex]s_1 =1.2[/tex] sample standard deviation for sample 1
[tex]s_2 =1.7[/tex] sample standard deviation for sample 2
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
Confidence interval
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_1 -\bar X_2 =5.1-6.3=-1.2[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n_1 +n_2 -1=250+280-2=528[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,528)".And we see that [tex]t_{\alpha/2}=1.964[/tex]
The standard error is given by the following formula:
[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex]
And replacing we have:
[tex]SE=\sqrt{\frac{1.2^2}{250}+\frac{1.7^2}{280}}=0.127[/tex]
Now we have everything in order to replace into formula (1):
[tex]-1.2-1.96\sqrt{\frac{1.2^2}{250}+\frac{1.7^2}{280}}=-1.4485[/tex]
[tex]-1.2+1.96\sqrt{\frac{1.2^2}{250}+\frac{1.7^2}{280}}=-0.9515[/tex]
So on this case the 95% confidence interval would be given by [tex]-1.4485 \leq \mu_G -\mu_B \leq -0.9515[/tex]
Since the confidence interval not contains the 0 we can say that we have significant differences between the mean of girls and boys.
1. What is the value of the lower end of the confidence interval?
b. – 1.4485
2. What is the value of the upper end of the confidence interval?
c. – 0.9515
What kind of distribution ( central limit theorem or T) and why?
We can use the t distribution but since the sample size is large enough we will have a distribution similar to the normal standard distribution. Because when the degrees of freedom of the t distribution increases we have a normal distribution.
What is the missing step for step 8
Answer:
CD ≅ CD, Reflexive property
Step-by-step explanation:
We want to show the triangles are similar by SAS. We've already proven that one pair of their sides are congruent (AC ≅ BD), and that one pair of their angles are congruent (∠CDE ≅ ∠DCE), so we need to show that the next pair of sides are congruent.
From the two column proof below, we have seen the missing step 8 is: CD ≅ CD, Reflexive property
How to solve two column proof problems?The two column proof to show that ΔACD ≅ ΔBCD
We are given that AC ≅ BD because congruent segments added to congruent segments form congruent segments.
We are also given that ∠CDE ≅ ∠DCE. This means we have one side and the included angle as congruent.
We need one more congruent side to prove Congruency. Thus:
CD ≅ CD, Reflexive property
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(2 points) The director of an alumni association for a university wants to look at the relationship between the number of years since graduation and the amount of monetary contribution an alumnus makes to the university. He collects data on 50 alumni who have made contributions this year and fits a least squares regression line to the data, with the monetary contribution as the response variable. James, one of the 50 alumni, has made a contribution which gives a negative residual. Which of the following statements must be true about James' actual contribution?
(A) It is less than the contribution predicted by the regression line.
(B) It is less than the average contribution made by the 50 alumni.
(C) Both (a) and (b)
(D) Neither (a) nor (b).
Answer:
(A) It is less than the contribution predicted by the regression line.
Step-by-step explanation:
The residual is the subtraction of the observed value by the predicted value.
So, if James has a negative residual, it means that his contribution is less than what was expected by the regression line.
The correct answer is:
(A) It is less than the contribution predicted by the regression line.
James' actual contribution must be less than the contribution predicted by the regression line.
Explanation:To determine which statement must be true about James' actual contribution, we need to understand the concept of residuals in regression analysis. A residual is the difference between the actual observed value and the predicted value. Since James' residual is negative, it means that his actual contribution is less than the contribution predicted by the regression line. (A) It is less than the contribution predicted by the regression line.
However, we cannot determine if James' actual contribution is less than the average contribution made by the 50 alumni solely based on the fact that his residual is negative. The regression line only predicts individual contributions based on the number of years since graduation, and the average contribution is not directly related to the regression line. Therefore, (B) It is less than the average contribution made by the 50 alumni. cannot be concluded.
Therefore, the correct answer is (A) It is less than the contribution predicted by the regression line.
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An article in the November 1983 Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline AA batteries and Eveready Energizer Alkaline AA batteries were given as 4.1 hours and 4.5 hours, respectively. Suppose these are the population average lifetimes. (a) Let X be the sample average lifetime of 81 Duracell and Y be the sample average lifetime of 81 Eveready Energizer batteries.
What is the mean value of X − Y (i.e., where is the distribution of X − Y centered)?
Does your answer depend on the specified sample sizes?
A. The answer increases as the sample size decreases.
B. The answer is the same irrespective of the sample sizes.
C. The answer decreases as the sample size increases.
D. The answer decreases as the sample size decreases.
E. The answer increases as the sample size increases.
Answer:
-0.4
B. The answer is the same irrespective of the sample sizes.
Step-by-step explanation:
Given that an article in the November 1983 Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline AA batteries and Eveready Energizer Alkaline AA batteries were given as 4.1 hours and 4.5 hours, respectively
Let X be the sample average lifetime of 81 Duracell and Y be the sample average lifetime of 81 Eveready Energizer batteries.
Now we have sample mean difference does not depend about the sample sizes.
This is because
E(X-Y) = E(X)-E(Y) for all X and Y
Mean of X-Y = [tex]4.1-4.5=-0.4[/tex]
This does not depend on the specified sample sizes
B. The answer is the same irrespective of the sample sizes.
The mean value of X - Y is 0.4 hours. The answer does not depend on the sample sizes.
Explanation:The mean value of X - Y, where X is the sample average lifetime of 81 Duracell batteries and Y is the sample average lifetime of 81 Eveready Energizer batteries, is given by the difference in the population average lifetimes of the two types of batteries. In this case, the difference is 4.5 - 4.1 = 0.4 hours. So the distribution of X - Y is centered around 0.4 hours.
The answer does not depend on the specified sample sizes in this case. The mean value of X - Y remains the same regardless of the sample sizes, as long as the samples are representative of the populations and the population parameters do not change. So the answer is B. The answer is the same irrespective of the sample sizes.
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You are driving on an asphalt road that had a 40 mi/h speed limit. A bicyclist veered into your lane so you slam on your brakes. Your tires left three skid marks of 69 ft, 70 ft, and 74 ft. The road had a drag factor of 0.95. Your brakes were operating at 98% efficiency. The police gave you a ticket for speeding. You insist that you were driving under the speed limit. Who is correct?
Answer:
Policeman is correct.
Step-by-step explanation:
Consider the provided information.
The tires left three skid marks of 69 ft, 70 ft, and 74 ft.
The road had a drag factor of 0.95. Your brakes were operating at 98% efficiency.
Thus, drag factor (f) = 0.95 and brakes efficiency (n) = 0.98
Compute the skid distance by finding the mean of length of 3 skids.
[tex]D=\frac{69+70+74}{3}\\D=\frac{213}{3}\\D=71[/tex]
Therefore, the skid distance is 71 ft.
Now find the speed of car using skid speed formula:
[tex]S=\sqrt{30Dfn}[/tex]
Where, S is the speed of car, D is the skid distance, f is the drag factor and n is the beak efficiency.
Substitute the respective values in above formula.
[tex]S=\sqrt{30\times 71\times0.95 \times0.98}[/tex]
[tex]S=\sqrt{1983.03}[/tex]
[tex]S\approx 44.5312[/tex]
Your speed was 44.5312 mi/h which is greater than 40 mi/hr. So policeman is correct.
Mel needs to measure 4 3/4 cups of flour. The only measuring cup he has measures 1/4 cup. How many 1/4 cups of flour should he use?
Answer:
19
Step-by-step explanation:
Given:
measuring cup size = 1/4 cup
total amount to be measured,
= 4 3/4 cups (convert to improper fraction)
= 19/4 cups.
number of 1/4 cups
= total amount measured ÷ 1/4 cup
= 19/4 ÷ 1/4
= 19/4 x 4/1
= 19/4 x 4
= 19 (1/4-cups) used
To figure out how many 1/4 cups Mel will need, we need to convert 4 3/4 cups to quarters. As 4 cups equal 16 quarters and 3/4 equals 3 quarters, we have a sum of 19 quarters. Therefore, Mel will need 19 measures of his 1/4 cup.
Explanation:To solve this problem, you need to understand how to convert measurements using fractions. We know that Mel needs 4 3/4 cups of flour and the only measuring cup he has measures 1/4 cup.
The first thing we can do is convert 4 3/4 into an improper fraction. One whole cup is 4 quarters, so 4 cups are 4 x 4 = 16 quarters. Additionally, there are 3 more quarter cups, totaling 16 + 3 = 19 quarter cups.
So, Mel needs 19 quarter cups of flour.
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A car company has found that there is a linear relationship between the money it spends on advertising and the number of cars it sells. When it spent 60000 dollars on advertising, it sold 680 cars. Moreover, for each additional 5000 dollars spent, they sold 40 more cars. Let x be the amount money they spend on advertising, in thousands of dollars. Find a formula for y, the number of cars sold.
Answer:
The required formula for y, the number of cars sold is: [tex]y = 0.008x +200[/tex].
Step-by-step explanation:
Consider the provided information.
Let x represents the amount of money spend on advertising .
y is number of cars sold,
For each additional 5000 dollars spent, they sold 40 more cars.
[tex]Slope = m = \frac{Rise}{Run}=\frac{40}{5000}=0.008[/tex]
The y-intercept form is: [tex]y = mx + b[/tex]
Substitute x=60,000, y=680 and m=0.008 in the above equation.
[tex]680= 0.008(60,000)+b[/tex]
[tex]b=680-480[/tex]
[tex]b=200[/tex]
Thus, the required formula for y, the number of cars sold is: [tex]y = 0.008x +200[/tex].
To create a formula for the number of cars sold (y) as a function of advertising expenditure (x, in thousands of dollars), we use the data points provided to establish a linear equation: y = 8x + 200, where x represents the advertising spending and y the number of cars sold.
To find a formula for y, the number of cars sold, given the linear relationship between advertising expenditure (x, in thousands of dollars) and sales, we first establish the given data points:
An additional $5,000 in advertising (or an increase of 5 in x since it's measured in thousands) results in selling 40 more cars.
We can express this relationship with the linear equation y = mx + b, where m is the slope (change in y over change in x) and b is the y-intercept (the number of cars sold when no money is spent on advertising).
Let's calculate the slope, m:
For every increase of 5 in x, y increases by 40, so m = 40/5 = 8.
We can then use this slope and one of the data points to find the y-intercept, b:
680 = 8(60) + b, so b = 680 - (8*60) = 680 - 480 = 200.
The linear equation representing the relationship between advertising spend and cars sold is:
y = 8x + 200
Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis.
y = 2x, y = 2, x = 0
Answer:
V = 8π/3
Step-by-step explanation:
Haven't done calc in years so I might be wrong.
Graph out all the lines needed so you can have a better look at it.
I set y=2x = y=2 to find where the intersect each others so I can have my boundaries for integration.
You goal is to find the area so you can integrate around that area. We're revolving around the x-axis so the area will be a circle.
V = ∫A(x)dx = ∫(πr²)dr
Since we have two different radius, we subtract them from each others.
∫(πr₂² - πr₁²)dr
∫(π(2)² - π(2x)²)dr
∫(4π - 4πx²)dr
4π∫(1 - x²)dr
integrate from 0 to 1 since that's where our boundary is.
V = 4π∫(1 - x²)dr = 8π/3
To find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis, we can use the method of cylindrical shells and integrate the expression 2πx(2-2x) from 0 to 1.
Explanation:To find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis, we can use the method of cylindrical shells. The region is bounded by the line y = 2x, the horizontal line y = 2, and the vertical line x = 0. First, we need to determine the limits of integration by finding the points where the curves intersect. Setting y = 2x and y = 2 equal to each other, we find x = 1. Next, we integrate the expression 2πx(2-2x) with respect to x from 0 to 1. Simplifying and evaluating the integral gives us the volume of the solid generated as 2π/3 cubic units.
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Ted owns a small florist shop. Since his business is booming, his realizes he will soon need one more delivery van. He decides he will purchase a full size van versus a minivan, which he currently owns. The van he is looking to buy in 3 years will cost him $25,000. How much should he invest each quarter into an account that pays 3% per year compounded quarterly, so that he can have the desired funds in 3 years?
a) Present Value with compound interestb) Sinking Fundc) Amortizationd) Present Value of an Annuitye) Future Value with compound interestf) None of the above.
Answer:The type of the problem described above is a Sinking Fund
Option B
Step-by-step explanation:
In order to understand the solution to this question we have to be familiar with these concepts
Sinking Fund
A sinking fund is an account earning compound interest into which you make periodic deposits. Suppose that the account has an annual interest rate of (r) compounded (m) times per year, so that (i=r/m) is the interest rate per compounding period. If you make a payment of PMT at the end of each period, then the future value after (t) years, or (n = mt) periods, will be
FV = [tex] PMT (〖(1+i )〗^n -1)/i [/tex]
Where FV is the amount that would be accumulated after t years
Payment Formula for a Sinking Fund
Suppose that an account has an annual rate of (r) compounded (m) times per year, so that is (i=r/m) is the interest rate per compounding period. If you want to accumulate a total of FV in the account after t years, or (n = mt) periods, by making payments of PMT at the end of each period, then each payment must be
PMT = [tex] FV ( i)/(〖(1+i)〗^n -1) [/tex]
From the question
Rate= r = 3/100 = 0.03
Number of times it was paid (compounded) in a year = m = 4 its value is Four cause the payment is made 4 times in one year i.e. Quarterly
The interest rate per compounding period = I = r/m = 0.03/4 = 0.0075
Number of times it was paid (compounded) t years n = 4 x 3 = 12
The amount that ted desires to be in that account after 3 years =FV = $25,000
So the investment that Ted needs to make Quarterly in order to get his desired amount is
= [tex]25000 × (0.0075/(〖(1+0.0075)〗^12 -1 )) [/tex]
= $2000
The correct option is b. Sinking Fund. Ted should invest approximately $22,857.14 each quarter to have $25,000 in 3 years with an annual interest rate of 3% compounded quarterly.
To determine how much Ted should invest each quarter into an account that pays 3% per year compounded quarterly, we can use the sinking fund formula:[tex]\[ P = \frac{FV}{\left(1 + \frac{r}{m}\right)^{n \cdot m}} \][/tex]
where:
P is the payment made each period (quarterly in this case),
FV is the future value of the investment, which is $25,000,
r is the annual interest rate (3% or 0.03),
m is the number of times interest is compounded per year (4 for quarterly),
n is the number of years (3).
First, we need to adjust the interest rate for quarterly compounding:
[tex]\[ \text{Quarterly interest rate} = \frac{r}{m} = \frac{0.03}{4} = 0.0075 \][/tex]
Next, we calculate the number of total compounding periods:
[tex]\[ n \cdot m = 3 \cdot 4 = 12 \][/tex]
Now we can plug these values into the sinking fund formula to solve for P:
[tex]\[ P = \frac{25000}{\left(1 + 0.0075\right)^{12}} \][/tex]
[tex]\[ P = \frac{25000}{\left(1.0075\right)^{12}} \][/tex]
[tex]\[ P = \frac{25000}{1.0934} \][/tex]
[tex]\[ P \approx \frac{25000}{1.0934} \][/tex]
[tex]\[ P \approx 2285.71 \][/tex]
Therefore, Ted should invest approximately $22,857.14 each quarter to have $25,000 in 3 years with an annual interest rate of 3% compounded quarterly.
A study is going to be conducted in which a mean of a lifetime of batteries produced by a certain method will be estimated using a 90% confidence interval. The estimate needs to be within +/- 2 hours of the actual population mean. The population standard deviation s is estimated to be around 25. The necessary sample size should be at least _______.
Answer:
The necessary sample size should be at least 423.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the length of the sample.
In this problem, we have that:
[tex]M = 2, \sigma = 25[/tex]. So
[tex]2 = 1.645*\frac{25}{\sqrt{n}}[/tex]
[tex]2\sqrt{n} = 41.125[/tex]
[tex]\sqrt{n} = 20.5625[/tex]
[tex]\sqrt{n}^{2} = (20.5625)^{2}[/tex]
[tex]n = 422.81[/tex]
The necessary sample size should be at least 423.
An individual is teaching a class on Excel Macros. The individual plans to break the class up into groups of 4 and wants each group to have 2 exercises to practice on, with no group doing the same exercise. The individual wants to know how many exercises he will need. Write an equation that expresses the situation, let x be the independent variable and y be the dependent variable
1. У-4x
2. y-2(x/4)
3. y-4x/2
Answer:
Total Exercises each group will do =[tex]2\frac{x}{4}[/tex]
Total exercises individual need= [tex]y=\frac{4x}{2}[/tex]
[tex]y-\frac{4x}{2}[/tex]
Step-by-step explanation:
No. of groups = 4
Each group has to do exercises = 2
Total no. of exercises individual need = y
Total Exercises each group will do =[tex]2\frac{x}{4}[/tex]
Total exercises individual need= [tex]y=2\frac{x}{4}(4)[/tex]
[tex]y=\frac{4x}{2}[/tex]
[tex]y-\frac{4x}{2}[/tex]
Answer:
y-2{x/4}
Step-by-step explanation:
A certain transportation system of buses and commuter trains is heavily utilized so that it is not practical to check every traveler's ticket. Rather, only a small, randomly selected group of travelers on any given trip will be asked to show their tickets. Suppose that in a random sample of 621 train travelers is selected and 69 of them admitted they did not buy a ticket. Find the upper bound of a 95% confidence interval for the true proportion of all train travelers who do not buy tickets
Answer:
[tex]0.0864 < p< 0.1358[/tex]
We are confident (95%) that the true proportion of people admitted they did not buy a ticket is betwen 0.0864 and 0.1358.
Step-by-step explanation:
1) Data given and notation
n=621 represent the random sample taken
X=69 represent the people admitted they did not buy a ticket
[tex]\hat p=\frac{69}{621}=0.111[/tex] estimated proportion of people admitted they did not buy a ticket
[tex]\alpha=0.05[/tex] represent the significance level (no given, but is assumed)
z would represent the statistic
p= population proportion of people admitted they did not buy a ticket
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
2) Confidence interval
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.111 - 1.96 \sqrt{\frac{0.111(1-0.111)}{621}}=0.0864[/tex]
[tex]0.111 + 1.96 \sqrt{\frac{0.111(1-0.111)}{621}}=0.1358[/tex]
And the 95% confidence interval would be given (0.0864;0.1358).
[tex]0.0864 < p< 0.1358[/tex]
We are confident (95%) that the true proportion of people admitted they did not buy a ticket is betwen 0.0864 and 0.1358.
The table below gives the number of hours ten randomly selected students spent studying and their corresponding midterm exam grades. Using this data, consider the equation of the regression line, y = bo + b1x, for predicting the midterm exam grade that a student will earn based the number of hours spent studying. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.
Hours Studying 0 1 1.5 2 2.5 3 4.5 5 5.5 6
Midterm Grades 65 70 77 79 83 91 92 94 95 98
a. Find the estimated slope. Round your answer to three decimal places.
b. Find the estimated y-intercept. Round your answer to three decimal places.
c. Find the estimated value of y when x=5. Round your answer to three decimal places.
d. Find the error prediction when x=2. Round your answer to three decimal places.
The estimated slope is approximately -4.028 and the estimated y-intercept is approximately 92.919.
The estimated value of y when x=5 is approximately 72.779, and the error prediction when x=2 is approximately 5.863.
The estimated slope (b1) and the estimated y-intercept (bo) in the regression equation y = bo + b1x, you can use the following formulas:
[tex]\[b1 = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}\][/tex]
[tex]\[bo = \frac{\sum y - b1(\sum x)}{n}\][/tex]
Where:
n is the number of data points,
[tex]\(\sum xy\)[/tex] is the sum of the product of x and y,
[tex]\(\sum x\)[/tex] is the sum of x,
[tex]\(\sum y\)[/tex] is the sum of y,
[tex]\(\sum x^2\)[/tex] is the sum of the squared x values.
Now let's calculate these values step by step:
a. Find the estimated slope (b1):
[tex]\[n = 10\][/tex]
[tex]\[\sum x = 31\][/tex]
[tex]\[\sum y = 804\][/tex]
[tex]\[\sum xy = 2439\][/tex]
[tex]\[\sum x^2 = 121.25\][/tex]
[tex]\[b1 = \frac{(10 \times 2439) - (31 \times 804)}{(10 \times 121.25) - 31^2} \][/tex]
[tex]\[b1 = \frac{24390 - 25404}{1212.5 - 961}\][/tex]
[tex]\[b1 = \frac{-1014}{251.5}\][/tex]
[tex]\[b1 \approx -4.028 \ (rounded to three decimal places)\][/tex]
b. Find the estimated y-intercept (bo):
[tex]\[bo = \frac{804 - (-4.028 \times 31)}{10}\][/tex]
[tex]\[bo = \frac{804 + 125.188}{10}\][/tex]
[tex]\[bo \approx \frac{929.188}{10}\][/tex]
[tex]\[bo \approx 92.919 \ (rounded to three decimal places)\][/tex]
c. Find the estimated value of y when x=5:
[tex]\[y = bo + b1x\][/tex]
[tex]\[y = 92.919 + (-4.028 \times 5)\][/tex]
[tex]\[y = 92.919 - 20.14\][/tex]
[tex]\[y \approx 72.779 \ (rounded to three decimal places)\][/tex]
d. Find the error prediction when x=2:
First, find the predicted y using the regression line:
[tex]\[y = bo + b1x\][/tex]
[tex]\[y = 92.919 + (-4.028 \times 2)\][/tex]
[tex]\[y = 92.919 - 8.056\][/tex]
[tex]\[y \approx 84.863 \ (rounded to three decimal places)\][/tex]
Now, find the error prediction:
[tex]\[Error = |Observed\, y - Predicted\, y|\][/tex]
[tex]\[Error = |79 - 84.863|\][/tex]
[tex]\[Error \approx 5.863 \ (rounded to three decimal places)\][/tex]
Annual salary plus bonus data for chief executive officers are presented in the BusinessWeek Annual Pay Survey. A preliminary sample showed that the standard deviation is $675 with data provided in thousands of dollars. How many chief executive officers should be in a sample if we want to estimate the population mean annual salary plus bonus with a margin of error of $100,000?
Answer: 176.
Step-by-step explanation:
Formula to find the sample size is given by :-
[tex]n=(\dfrac{z^*\cdot \sigma}{E})^2[/tex]
, where [tex]\sigma[/tex] = Population standard deviation from prior study.
E = margin of error.
z* = Critical value.
As per given , we have
[tex]\sigma=\$675000[/tex]
E= $100,000
We take 95% confidence interval.
Critical value (Two tailed)=[tex]z^*=1.96[/tex]
The required sample size = [tex]n=(\dfrac{(1.96)\cdot 675000}{100000})^2[/tex]
[tex]n=(13.23)^2\\\\ n=175.03299\approx176[/tex] [Round to next integer]
Hence, the required sample size = 176.
What is the solution to the equation shown below?
A.
x = 4.5
B.
x = 1.5
C.
x = 9
D.
x = 3
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-6 from both sides
3/x-2 = √(x - 2) + 2
Multiply both sides by (x - 2)
(Note: this cancels out the square root)
3 = x - 2 + 2
x = 3
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In a survey of 1016 ?adults, a polling agency? asked, "When you? retire, do you think you will have enough money to live comfortably or not. Of the 1016 ?surveyed, 535 stated that they were worried about having enough money to live comfortably in retirement. Construct a 99?% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.
A. There is a 99?% probability that the true proportion of worried adults is between ___ and ___.
B. 99?% of the population lies in the interval between ___ and ___.
C. There is 99?% confidence that the proportion of worried adults is between ___ and ___.
Answer:
C. There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567
Step-by-step explanation:
1) Data given and notation
n=1016 represent the random sample taken
X=535 represent the people stated that they were worried about having enough money to live comfortably in retirement
[tex]\hat p=\frac{535}{1016}=0.527[/tex] estimated proportion of people stated that they were worried about having enough money to live comfortably in retirement
[tex]\alpha=0.01[/tex] represent the significance level
Confidence =0.99 or 99%
z would represent the statistic
p= population proportion of people stated that they were worried about having enough money to live comfortably in retirement
2) Confidence interval
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
And replacing into the confidence interval formula we got:
[tex]0.527 - 2.58 \sqrt{\frac{0.527(1-0.527)}{1016}}=0.487[/tex]
[tex]0.527 + 2.58 \sqrt{\frac{0.527(1-0.527)}{1016}}=0.567[/tex]
And the 99% confidence interval would be given (0.487;0.567).
There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567
To build a 99% confidence interval, we first calculate our sample proportion by dividing the number of such instances by the total sample size. Next, we determine the standard error of the proportion, then our margin of error by multiplying the standard error by the Z value of the selected confidence level. Lastly, we determine the confidence interval by adding and subtracting the margin of error from the sample proportion.
Explanation:To construct a 99% confidence interval for the proportion of adults worried about having enough money to live comfortably in retirement, we will utilize statistical methods and proportions. First, we must calculate the sample proportion. The sample proportion (p) is equal to 535 (the number who are worried) divided by 1016 (the total number of adults surveyed).
Then, we find the standard error of the proportion which we get by multiplying the square root of ((p*(1-p))/n) where n is the number of adults sampled. The margin of error is found using the Z value corresponding to the desired confidence level, in this case, 99%. Multiply the standard error by this Z value. Lastly, we construct the confidence interval by taking the sample proportion (p) ± the margin of error.
The result will give you the 99% confidence interval - meaning we are 99% confident that the true proportion of adults who are worried about having enough money to live comfortably in retirement lies within this interval.
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The mean hourly wage for employees in goods-producing industries is currently $24.57 (Bureau of Labor Statistics website, April, 12, 2012). Suppose we take a sample of employees from the manufacturing industry to see if the mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries.
To determine if the mean hourly wage of employees in the manufacturing industry differs from the reported mean for the goods-producing industries, a hypothesis test can be conducted. Steps include setting up null and alternative hypotheses, selecting a significance level, collecting a sample, calculating a test statistic, determining critical values or p-values, and making a conclusion based on the results.
Explanation:To determine if the mean hourly wage of employees in the manufacturing industry differs from the reported mean of $24.57 for the goods-producing industries, we can conduct a hypothesis test.
Step 1: Set up the null and alternative hypotheses. The null hypothesis (H0) states that the mean hourly wage of employees in the manufacturing industry is equal to $24.57, while the alternative hypothesis (Ha) states that it is not equal to $24.57.Step 2: Select a significance level. This determines the threshold for rejecting the null hypothesis. Let's say we choose a significance level of 0.05 (5%).Step 3: Collect a sample of employees from the manufacturing industry and calculate their mean hourly wage.Step 4: Calculate the test statistic. In this case, we can use a t-test since we have sample data and want to compare it to a known population mean.Step 5: Determine the critical value(s) or p-value associated with the test statistic. If the test statistic falls in the rejection region (beyond the critical value(s)) or if the p-value is less than the significance level, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.Step 6: Make a conclusion based on the results. If we reject the null hypothesis, it suggests that the mean hourly wage of employees in the manufacturing industry differs from $24.57. If we fail to reject the null hypothesis, it suggests that there is not enough evidence to conclude a difference.Learn more about Hypothesis test here:https://brainly.com/question/34171008
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When a production process is operating correctly, thenumber of units produced per hour has a normal distri- bution with a mean of 92.0 and a standard deviation of3.6. A random sample of 4 different hours was taken.a. Find the mean of the sampling distribution of thesample means. b. Find the variance of the sampling distribution ofthe sample mean.c. Find the standard error of the sampling distribu-tion of the sample mean.d. What is the probability that the sample mean ex-ceeds 93.0 units?
Answer:
a) The mean of the sampling distribution of the sample means is 92.
b) The variance of the sampling distribution of the sample mean is 3.24.
c) The standard error of the sampling distribution of the sample mean is 1.8.
d) 28.77% probability that the sample mean ex-ceeds 93.0 units.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 92, \sigma = 3.6[/tex].
a. Find the mean of the sampling distribution of thesample means.
The mean is the same as the mean of the population. So the mean of the sampling distribution of the sample means is 92.
b. Find the variance of the sampling distribution ofthe sample mean.
The standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{4}} = 1.8[/tex]
The variance is [tex]s^{2} = 1.8^{2} = 3.24[/tex]
c. Find the standard error of the sampling distribution of the sample mean.
This is the same as the standard deviation of the sample. So the standard error of the sampling distribution of the sample mean is 1.8.
d. What is the probability that the sample mean ex-ceeds 93.0 units?
This is 1 subtracted by the pvalue of Z when X = 93. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{93 - 92}{1.8}[/tex]
[tex]Z = 0.56[/tex]
[tex]Z = 0.56[/tex] has a pvalue of 0.7123.
So there is a 1-0.7123 = 0.2877 = 28.77% probability that the sample mean ex-ceeds 93.0 units.
The mean of the sampling distribution of the sample means is 92.0. The variance is 0.81. The standard error is 0.9.
Explanation:a. The mean of the sampling distribution of the sample means can be calculated by finding the mean of the original population, which is 92.0.
b. The variance of the sampling distribution of the sample mean can be calculated using the formula (standard deviation/original population size)^2. In this case, it is (3.6/4)^2 = 0.81.
c. The standard error of the sampling distribution of the sample mean can be calculated by taking the square root of the variance. So, in this case, it is √0.81 = 0.9.
d. To find the probability that the sample mean exceeds 93.0 units, you can calculate the z-score using the formula (sample mean - population mean)/standard error. Then, you can find the probability of the z-score using a standard normal distribution table or calculator.
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A linear regression model is fitted to the data x y 37.0 65.0 36.4 67.2 35.8 70.3 34.3 71.9 33.7 73.8 32.1 75.7 31.5 77.9 with x as the input variable and y as the output variable. Find βˆ0, βˆ1, and ˆσ2. Construct a 99% confidence interval for the expected value of the output variable when the input variable is equal to 35.
Answer:
b0= 144.59
b= -2.12
Se²= 1.02
99%CI E(Y/X=35): [68.78; 71.99]
Step-by-step explanation:
Hello!
I've arranged the given data:
X: 37.0, 36.4, 35.8, 34.3, 33.7, 32.1, 31.5
Y: 65.0, 67.2, 70.3, 71.9, 73.8, 75.7, 77.9
The equation of the linear regression model is:
Yi= β₀ + βXi + εi
Where
Yi is the dependent variable
Xi is the independent variable
εi represents the errors or residues
β₀ is the intercept of the line
β is the slope
The conditions to make a linear regression analysis are:
For each given value of X, there is a population of Y~N(μy;σy²)
Each value of Y is independent of the others.
The population variances of each population of Y are equal.
From these conditions the following characteristic is deduced:
εi~N(0;σ²)
The parameters of the regression are:
β₀, β, and σ²
If the conditions are met then you can estimate the regression line:
Yi= bo * bXi + ei.
And the point estimation of the parameters can be calculated using the formulas:
β₀ ⇒ b0= (∑y/n)-b(∑x/n)
β ⇒ b= [∑xy- ((∑x)(∑y))/n]/(∑x²-((∑x)²/n))
σ²⇒ Se²= 1/(n-2)*[∑y²-(∑y)²/n - b²(∑x²-(∑x)²/n)]
n= 7
∑y= 501.80
∑y²= 36097.88
∑x= 240.80
∑x²= 8310.44
∑xy= 17204.87
b0= 144.59
b= -2.12
Se²= 1.02
The estimated regression line is:
Yi= 144.59 -2.12Xi
You need to calculate a 99%CI E(Y/X=35), the formula is:
(b0 + bX0) ± [tex]t_{n-2;1-\alpha /2}[/tex]*[tex]\sqrt{S_e^2(\frac{1}{n}+\frac{(X_0-X[bar])^2}{sumX^2-(\frac{(sumX)^2}{n} )} )}[/tex]
(144.59 + (-2.12*35)) ± 4.032*[tex]\sqrt{1.02(\frac{1}{7}+\frac{(35-34.4)^2}{8310.44-(\frac{(240.80)^2}{7} )} )}[/tex]
[68.78; 71.99]
With a 99% confidence level youd expect that the interval [68.78; 71.99] contains the true value of the average of Y when X= 35.
I hope it helps!
What is the slope and the y-intercept of the line on the graph below? Answer / a. Slope = -4, y-intercept = 1 B. Slope = -4, y-intercept =4 C. Slope = -1/4, y-intercept = 1 D. Slope = -1/4, y-intercept = 4
Answer: The correct answer is C.
Step-by-step explanation: The y-intercept is the point on the graph at which the line crosses over the y-axis. In this case, your y-intercept is 1, because the line crosses over the y-axis at 1. The slope formula is rise/run, or y2 -y1/x2-x1. However since you were not supplies with any coordinates, you'll have to go with rise (up/down) over run (left/right). The slope of the line falls one cube down ward, making the slope negative 1. The slope "runs" to the right 4 cubes, making the slope -1 (falls 1 cube)/4 (moves to the left/right 4 cubes).
Which equation represents the line that passes through (-6, 7) and (-3, 6)
Answer:
x + 3y - 16 = 0
Step-by-step explanation:
When two points, say [tex]$ (x_1, y_1) $[/tex] and [tex]$ (x_2, y_2) $[/tex] are given, the equation is determined using Two - point form.
The two - point form is as follows:
[tex]$ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} $[/tex]
Here: [tex]$ (x_1, y_1) = (-6, 7) $[/tex] and [tex]$(x_2, y_2) = (-3, 6) $[/tex].
Substituting in the formula we get the equation of the line as:
[tex]$ \frac{y - 7}{6 - 7} = \frac{x + 6}{- 3 + 6}[/tex]
[tex]$ \implies \frac{y - 7}{- 1} = \frac{x + 6}{ 3} $[/tex]
[tex]$ \implies 3y - 21 = - x - 6 $[/tex]
Rearranging we get: x + 3y - 15 = 0
This is the required equation of the line.
A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 respondents, 12 % chose chocolate pie, and the margin of error was given as plus or minus 5 percentage points. What values do ModifyingAbove p with caret , ModifyingAbove q with caret , n, E, and p represent? If the confidence level is 90 %, what is the value of alpha ?
Answer:
n=500 represent the random sample taken
[tex]\hat p=0.12[/tex] estimated proportion of people that chose chocolate pie
[tex]\hat q =1-\hat p=1-0.12=0.88[/tex] represent the people that NOT chose chocolate pie
E=0.05 represent the error or margin of error given by the following formula:
[tex]ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
p= true population proportion of people that chose chocolate pie
If the confidence level is 90 %, what is the value of alpha ?
[tex]\alpha=1-0.9 =0.1[/tex] and the value of [tex]\alpha/2 =0.05[/tex],
[tex]z_{\alpha/2}=-1.64[/tex] and [tex]z_{1-\alpha/2}=1.64[/tex]
[tex]ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238[/tex]
Step-by-step explanation:
Data given and notation
What values do ModifyingAbove p with caret , ModifyingAbove q with caret , n, E, and p represent?
n=500 represent the random sample taken
X represent the people that chose chocolate pie
[tex]\hat p=0.12[/tex] estimated proportion of people that chose chocolate pie
[tex]\hat q =1-\hat p=1-0.12=0.88[/tex] represent the people that NOT chose chocolate pie
E=0.05 represent the error or margin of error given by the following formula:
[tex]ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
z would represent the quantile of the normal standard distribution
p= true population proportion of people that chose chocolate pie
The confidence interval for the population proportion is given by this formula :
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
If the confidence level is 90 %, what is the value of alpha ?
On this case the value for the significance would be [tex]\alpha=1-0.9 =0.1[/tex] and the value of [tex]\alpha/2 =0.05[/tex], we can find the quantiles of the normal standard distribution given by:
[tex]z_{\alpha/2}=-1.64[/tex] and [tex]z_{1-\alpha/2}=1.64[/tex]
And with the following excel codes:
"=NORM.INV(0.05,0,1)" "=NORM.INV(1-0.05,0,1)"
And we can find the margin of error like this:
[tex]ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238[/tex]
Final answer:
In the poll,
ModifyingAbove p with caret and
ModifyingAbove q with caret represent the sample proportions of respondents who chose and did not choose chocolate pie, respectively. The symbol n represents the sample size, E denotes the margin of error, and p is the population proportion. If the confidence level is 90%, the value of alpha is 0.10.
Explanation:
In the context of the provided poll scenario, the various symbols represent the following statistical terms:
ModifyingAbove p with caret (
ModifyingAbove p with caret) represents the sample proportion, which is the observed percentage in the sample that chose chocolate pie. In this case, it would be 0.12 or 12%.
ModifyingAbove q with caret (
ModifyingAbove q with caret) is the sample proportion of respondents who did not choose chocolate pie, which would be 1 - 0.12 = 0.88 or 88%.
n represents the sample size, which is the number of respondents in the poll, 500 in this scenario.
E denotes the margin of error, which is
plus or minus 5 percentage points in this case.
p represents the population proportion, which is the true percentage of all adults who would choose chocolate pie as their favorite if everyone was surveyed.
If the confidence level is 90%, the value of alpha (
alpha) is the probability that the true population parameter will not be contained in the confidence interval. For a 90% confidence level, alpha would be 1 - 0.90 = 0.10 or 10%. This is often split into two tails of the normal distribution for a two-tailed test, thus each tail would have an area of 0.05.
The Bernsteins, Hendersons, and Smiths each have five children. If the 15 children of these three families camp out in five different tents, where each tent holds three children, and the 15 children are randomly assigned to the five tents, what is the probability that every family has at least two of its children in the same tent?
Answer:
Probability=1-[tex]\frac{(5!)^{3} }{15!}[/tex]
Step-by-step explanation:
We have to solve this question with the help of complementary method. Observe the statement "Probability that every family has at least two of its children in the same tent". The complementary of this statement will be every family does not have atleast two of its children in the same tent ie. Each child of a family is not in the same tent as one from the same family.
P(A)=1-[tex]P(A)^{c}[/tex]
Therefore, we can get P(A) if we just take out the value of [tex]P(A)^{c}[/tex]
Probability=[tex]\frac{TotalNo.OfFavourableOutcomes}{TotalNo.ofOutcomes}[/tex]
Imagine that we have 15 locations to fill and 15 people for it, so the total no. of cases= 15!
Bernstein's children can be arranged in 5 different tents in 5! ways.
Similarly Henderson's and Smith's children can be arranged in 5 different tents in 5! ways only.
Therefore, [tex]P(A)^{c}[/tex]= [tex]\frac{(5!)^{3} }{15!}[/tex]
P(A)=1- [tex]\frac{(5!)^{3} }{15!}[/tex]
Definitions and Data Due Sun 02/03/2019 11:59 pm A researcher is interested in attitudes towards releasing prisoners with Alzheimer's who have a life sentence. 85 randomly selected Americans were asked, "Prisoners with Alzheimer's who have a life sentence should be released: Strongly Agree, Agree, Disagree, Strongly Disagree". Match the vocabulary word with its corresponding example. B All Americans The proportion of all Americans who strongly agree that prisoners with life sentences who have Alzheimer's should be released The proportion of the 85 Americans surveyed who strongly agree that prisoners with life sentences who have Alzheimer's should be released СВ The answer-Strongly Agree' or 'Agree, or Disagree, or Strongly Disagree, DThe 85 Americans who participated in the survey D The list of answers that the 85 Americans gave a. Data b. Variable c. Parameter d. Statistic e Sample f. Population 0 0
Answer:
Step-by-step explanation:
Matching each vocabulary word with its corresponding example:
(1). All Americans = f. Population (a group of items, units or subjects under reference of study e.g. inhabitants of a region, numbers of cars in a city, workers in a factory and so on)
(2). The proportion of all Americans who strongly agree that prisoners with life sentences who have Alzheimer's should be released = c. Parameter ( the number that summarizes some characteristic of a population)
(3). The proportion of the 85 Americans surveyed who strongly agree that prisoners with life sentences who have Alzheimer's should be released = d. Statistic (the sample characteristic corresponding to a population parameter used when a sample is used to make statistical inference about a population)
(4). The answer-Strongly Agree' or 'Agree, or Disagree, or Strongly Disagree = b. Variable (any quality, characteristic, quantity or number that can be counted or measured)
(5). The 85 Americans who participated in the survey = e. Sample (a part or fraction of a population selected on some basis)
(6). The list of answers that the 85 Americans gave = a. Data (the pieces of information collected to be used for the analysis)
Final answer:
In the research question provided, the key terms like Population, Parameter, Statistic, Sample, Variable, and Data are matched with examples based on a Likert-Response Scale survey on the release of prisoners with Alzheimer's who have life sentences.
Explanation:
A researcher inquiring about the public's view on releasing prisoners with Alzheimer's who have a life sentence used a Likert-Response Scale to understand the level of agreement with a statement. This method has several components we can define using the given options:
Population - (B) All Americans
Parameter - (C) The proportion of all Americans who strongly agree that prisoners with life sentences who have Alzheimer's should be released
Statistic - (D) The proportion of the 85 Americans surveyed who strongly agree that prisoners with life sentences who have Alzheimer's should be released
Variable - (CV) The answer 'Strongly Agree' or 'Agree, or Disagree, or Strongly Disagree'
Sample - (E) The 85 Americans who participated in the survey
Data - (F) The list of answers that the 85 Americans gave
In this study, the variable constitutes the different levels of agreement, while the data consists of the actual responses collected from the survey participants. The sample involves the subset of the population, which is the 85 Americans who were surveyed, and the statistic is the result derived from this sample. The population, on the other hand, includes all Americans, and the parameter is the value that would be obtained if all Americans could be surveyed.
The quality-control manager at a compact fluorescent light bulb (CFL) factory needs to determine whether the mean life of a large shipment of CFLs is equal to 7463 hours. The population standard deviation is 1080 hours. A random sample of 81 light bulbs indicates a sample mean life of 7163 hours.a. At the 0.05 level of significance, is there evidence that the mean life is different from 7 comma 463 hours question markb. Compute the p-value and interpret its meaning.c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs.d. Compare the results of (a) and (c). What conclusions do you reach?a. Let mu be the population mean. Determine the nullhypothesis, Upper H 0, and the alternative hypothesis, Upper H 1.Upper H 0: Upper H 1:What is the test statistic?Upper Z STAT (Round to two decimal places as needed.)What is/are the critical value(s)? (Round to two decimal places as needed. Use a comma to separate answers as needed.)What is the final conclusion?A. Reject Upper H 0. There is sufficient evidence to prove that the mean life is different from 7463 hours.B. Fail to reject Upper H 0. There is sufficient evidence to prove that the mean life is different from 7463 hours.C. Fail to reject Upper H 0. There is not sufficient evidence to prove that the mean life is different from 7463 hours.D. Reject Upper H 0. There is not sufficient evidence to prove that the mean life is different from 7463 hours.b. What is the p-value? (Round to three decimal places asneeded.)Interpret the meaning of the p-value. Choose the correct answer below.A. Fail to reject Upper H 0. There is not sufficient evidence to prove that the mean life is different from 7463 hours.B. Reject Upper H 0. There is sufficient evidence to prove that the mean life is different from 7463 hours.C. Reject Upper H 0. There is not sufficient evidence to prove that the mean life is different from 7463 hours.D. Fail to reject Upper H 0. There is sufficient evidence to prove that the mean life is different from 7463 hours.c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs. (Round to one decimal place as needed.)d. Compare the results of (a) and (c). What conclusions do you reach?A. The results of (a) and (c) are the same: there is not sufficient evidence to prove that the mean life is different from 7463 hours.B. The results of (a) and (c) are the same: there is sufficient evidence to prove that the mean life is different from 7463 hours.C. The results of (a) and (c) are not the same: there is sufficient evidence to prove that the mean life is different from 7463 hours.D. The results of (a) and (c) are not the same: there is not sufficient evidence to prove that the mean life is different from 7463 hours.
Answer:
Reject the null hypothesis. There is sufficient evidence to prove that the mean life is different from 7463 hours.
95% confidence interval also supports this result.
Step-by-step explanation:
Let mu be the population mean life of a large shipment of CFLs.
The hypotheses are:
[tex]H_{0}[/tex]: mu=7463 hours
[tex]H_{a}[/tex]: mu≠7463 hours
Test statistic can be calculated using the equation:
z=[tex]\frac{X-M}{\frac{s}{\sqrt{N} } }[/tex] where
X is the sample mean life of CFLs (7163 hours) M is the mean life assumed under null hypothesis. (7463 hours) s is the population standard deviation (1080 hours)N is the sample size (81)Then z=[tex]\frac{7163-7463}{\frac{1080}{\sqrt{81} } }[/tex] = -2.5
p-value is 0.0124, critical values at 0.05 significance are ±1.96
At the 0.05 level of significance, the the result is significant because 0.0124<0.05. There is significant evidence that mean life of light bulbs is different than 7463 hours.
95% Confidence Interval can be calculated using M±ME where
M is the sample mean life of a large shipment of CFLs (7163 hours)ME is the margin of error from the meanmargin of error (ME) from the mean can be calculated using the formula
ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where
z is the corresponding statistic in the 95% confidence level (1.96)s is the standard deviation of the sample (1080 hours)N is the sample size (81)Then ME=[tex]\frac{1.96*1080}{\sqrt{81} }[/tex] =235.2
Thus 95% confidence interval estimate of the population mean life of the light bulbs is 7163±235.2 hours. That is between 6927.8 and 7398.2 hours.