Answer:
Step-by-step explanation:
1) point slope form is:
y-y1=m(x-x1)
y1=8
x1=4
m=(slope)
to find m: y2-y1/x2-x1
hence
-2-8/2-4=-10/-2=5
m=5:
y-8=5(x+4)
Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. A sample of 60 day-shift workers showed that the mean number of units produced was 334, with a population standard deviation of 23. A sample of 68 night-shift workers showed that the mean number of units produced was 341, with a population standard deviation of 28 units.At the .10 significance level, is the number of units produced on the night shift larger?1. This is a (Click to select)twoone-tailed test.2. The decision rule is to reject H0: μd ≥ μn if z < . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)3. The test statistic is z = . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)4. What is your decision regarding H0?
Answer:
Step-by-step explanation:
Let the subscripts d and n represent day and night respectively
The null hypothesis is
H0 : μd ≥ μn
The alternative hypothesis is
H1 : μd < μn
it is a one-tailed and also a right left test because of the greater than symbol in the alternative hypothesis.
The decision rule is to reject H0: μd ≥ μn If 0.10 > p value
Since the population standard deviations are known, we would use the formula to determine the test statistic(z score)
z = (xd - xn)/√σd²/nd + σn²/nn
Where
xd and xn represents sample means for day and night respectively.
σd and σn represents population standard deviations for day and night respectively.
nd and nn represents number of samples
From the information given,
xd = 334
xn = 341
σd = 23
σ2 = 28
nd = 60
nn = 68
z = (334 - 341)/√23²/60 + 28²/68
= - 7/√20.34607843138
z = - 1.55
From the normal distribution table, the probability value corresponding to the z score is 0.061
Since the level of significance, 0.1 > 0.061, we would reject H0
Therefore, there is enough evidence to conclude that there are more units produced on the night shift than on the day shift.
Final answer:
Clark Heter is conducting a one-tailed test to compare the mean production of day and night shifts. The decision to reject the null hypothesis that day shift production is greater or equal to night shift production is based on a critical value of -1.28 linked to a significance level of 0.10. The test statistic is computed from the given means and standard deviations for both samples.
Explanation:
Conducting a Two-Sample Z-Test
Clark Heter wants to determine if more units are produced on the night shift than on the day shift. Given the samples:
- Day-shift (n=60): mean = 334, population standard deviation = 23.
- Night-shift (n=68): mean = 341, population standard deviation = 28.
The significance level is 0.10.
Teachers’ salaries in one state are very low that the educators in that state regularly complain about their compensation. The state mean is $33,600, but teachers in one district claim that the mean their district is significantly lower. They survey a simple random sample of 22 teachers in the district and calculate a mean salary of $32,400 with a standard deviation s = $ 1520. Test the teachers’ claim at the 0.05 level of significance.
Answer:
[tex]t=\frac{32400-33600}{\frac{1520}{\sqrt{22}}}=-3.702[/tex]
The degrees of freedom are given by:
[tex]df=n-1=22-1=21[/tex]
The p value is given by:
[tex]p_v =P(t_{(21)}<-3.702)=0.00066[/tex]
The p value is significantly lower than the significance level so then we have enough evidence to conclude that the true mean is significantly lower from 33600
Step-by-step explanation:
Information given
[tex]\bar X=33400[/tex] represent the sample mean
[tex]s=1520[/tex] represent the sample standard deviation
[tex]n=22[/tex] sample size
[tex]\mu_o =33600[/tex] represent the value that we want to analyze
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value for the test
System of hypothesis
We want to check if the true mean is lower than 33600, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 33600[/tex]
Alternative hypothesis:[tex]\mu < 33600[/tex]
The statistic is given:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the data given we got:
[tex]t=\frac{32400-33600}{\frac{1520}{\sqrt{22}}}=-3.702[/tex]
The degrees of freedom are given by:
[tex]df=n-1=22-1=21[/tex]
The p value is given by:
[tex]p_v =P(t_{(21)}<-3.702)=0.00066[/tex]
The p value is significantly lower than the significance level so then we have enough evidence to conclude that the true mean is significantly lower from 33600
Among 12 metal parts produced in a machine shop, 3 are defective.
Ok, what's the full question here
HELP
A family’s lunch bill is $10.19 before tax and tip. Using the percents shown for sales tax and gratuity, how much money should the family pay if the gratuity is calculated after tax?
$11.80
$12.16
$12.28
$12.36
Answer:
12.36
Step-by-step explanation:
An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 39 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?
Answer:
We conclude that the mean time taken to finish undergraduate degrees is longer than 4.5 years.
Step-by-step explanation:
We are given that an article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees.
You conduct a survey of 39 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2.
Let [tex]\mu[/tex] = average time taken to finish their undergraduate degrees.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 4.5 years {means that the mean time taken to finish undergraduate degrees is equal to 4.5 years}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 4.5 years {means that the mean time taken to finish undergraduate degrees is longer than 4.5 years}
The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean time = 5.1 years
s = sample standard deviation = 1.2 years
n = sample of students = 39
So, the test statistics = [tex]\frac{5.1-4.5}{\frac{1.2}{\sqrt{39} } }[/tex] ~ [tex]t_3_8[/tex]
= 3.122
The value of t test statistics is 3.122.
Now, at 1% significance level the t table gives critical value of 2.429 at 38 degree of freedom for right-tailed test.
Since our test statistic is more than the critical value of t as 3.122 > 2.429, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the mean time taken to finish undergraduate degrees is longer than 4.5 years.
Adrian earns $16000 per month. He spends 1/4 of his income on food; 3/10 of the remainder on house rent . How much money does he have left?
Answer:
8000
Step-by-step explanation:
1/4 of 16000 is 4000
16000-4000=12000 1/3 of 12000 is also 4000
12000-4000=8000
One kitty weighs 2 pounds 4 ounces. Another kitten weighs 2 ounces less. What is the combined weight of the two kittens in ounces
The combined weight of the two kittens is 70 ounces.
To find the combined weight of the two kittens, we'll start by converting the weight of the first kitten to ounces.
1 pound is equal to 16 ounces, so 2 pounds is equal to 2 x 16 = 32 ounces.
Therefore, the first kitten weighs 32 + 4 = 36 ounces.
The weight of the second kitten is 2 ounces less, so we subtract 2 from the weight of the first kitten: 36 - 2 = 34 ounces.
Finally, we can find the combined weight by adding the weights of the two kittens together: 36 + 34 = 70 ounces.
Therefore, the combined weight of the two kittens is 70 ounces.
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simplify the expression below. (-3x2 + 2x - 4) + (4x2 + 5x +9)
Answer: =x2+7x+5
Step-by-step explanation:
−3x2+2x−4+4x2+5x+9
=−3x2+2x+−4+4x2+5x+9
Combine Like Terms:
=−3x2+2x+−4+4x2+5x+9
=(−3x2+4x2)+(2x+5x)+(−4+9)
=x2+7x+5
Answer:
=x2+7x+5
Answer:
x^2+7x+5
Step-by-step explanation:
Remove parentheses: -3x^2+2x-4+4x^2+5x+9
Group like terms: -3x^2+4x^2+2x+5x-4+9
Add similar elements: x^2+2x+5x-4+9
Add/subtract the numbers: x^2+7x+5
Nana Akosua Owusu – Ansah, a financial manageress for a company is considering two competing investment proposals. For each of these proposals, she has carried out an analysis in which she has determined various net profit figures and has assigned subjective probabilities to the realization of these returns. For proposal A, her analysis shows net profits of GHȼ 20,000.00, GHȼ 30,000.00 or GHȼ 50,000.00 with respective probabilities 0.2, 0.4 and 0.4. For proposal B, she concludes that there is a 50% chance of successful investment, estimated as producing net profits of GHȼ 100,000.00, and of an unsuccessful investment, estimated as a break – even situation involving GHȼ 0.00 of net profit. Assuming that each proposal requires the same Ghana cedi investment, which of the two proposals is preferable solely from the standpoint of expected monetary return?
Answer:
Proposal B
Step-by-step explanation:
This problem can be solved by comparing the expected returns on both options.
The expected return is the sum of the possible outcomes multiplied by its probabilities of occurrence.
For proposal A, the net profits are $20,000, $30,000 and $50,000, with respective probabilities 0.2, 0.4 and 0.4. Then, the expected return can be calculated as:
[tex]E(A)=\sum_{i=1}^3p_iR_i\\\\E(A)=p_1R_1+p_2R_2+p_3R_3\\\\E(A)=0.2*20,000+0.4*30,000+0.4*50,000\\\\E(A)=4,000+12,000+20,000\\\\E(A)=36,000[/tex]
The proposal A has a expected net profit of $36,000.
The proposal B has a 50% chance of having a net profit of $100,000 and a 50% of break even (zero net profit). We applied the same calculation for the expected profit and we have:
[tex]E(B)=\sum_{i=1}^2p_iR_i\\\\E(B)=p_1R_1+p_2R_2\\\\E(B)=0.5*100,000+0.5*0\\\\E(B)=50,000[/tex]
The proposal B has a expected net profit of $50,000.
Assuming that each proposal requires the same investment, the proposal B has more expected monetary return (GHȼ 50,000) than proposal A (GHȼ 36,000).
(b) Based on the summary statistics, would it be more likely to obtain a yield
of 123 or more bushels per acre from a plot of GM corn or a plot of regular
corn? Justify your answer.
Answer: yes, GM is expected to provide performance greater than 123
Step-by-step explanation:
b) It is expected that the GM achieves a performance greater than 123. This situation occurs because 123 is less than the average for GM but is greater than the average for Regular. Thus we observe that P (performance> 123 | GM)> 0.5 and P (performance> 123 | Regular) <0.5
Can anyone pls help me to solve question 2 f and g and pls provide me a explanation I’m with that questions for three days
Answer:
f) a[n] = -(-2)^n +2^n
g) a[n] = (1/2)((-2)^-n +2^-n)
Step-by-step explanation:
Both of these problems are solved in the same way. The characteristic equation comes from ...
a[n] -k²·a[n-2] = 0
Using a[n] = r^n, we have ...
r^n -k²r^(n-2) = 0
r^(n-2)(r² -k²) = 0
r² -k² = 0
r = ±k
a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q
We find p and q from the initial conditions.
__
f) k² = 4, so k = 2.
a[0] = 0 = p + q
a[1] = 4 = -2p +2q
Dividing the second equation by 2 and adding the first, we have ...
2 = 2q
q = 1
p = -1
The solution is a[n] = -(-2)^n +2^n.
__
g) k² = 1/4, so k = 1/2.
a[0] = 1 = p + q
a[1] = 0 = -p/2 +q/2
Multiplying the first equation by 1/2 and adding the second, we get ...
1/2 = q
p = 1 -q = 1/2
Using k = 2^-1, we can write the solution as follows.
The solution is a[n] = (1/2)((-2)^-n +2^-n).
At the post office, Tiffany paid, $11.04. for 2 stamps. At this rate, how much would it cost for Tiffany yo buy 18 stamps?
Answer: $$$99.36
Step-by-step explanation:
It's 11.04 for 2 stamps
U have to buy 18 stamps
18÷2=9
9×11.04=99.36
Answer:
It costs $5.52 to Sam to buy 12 stamps.now multiply
Step-by-step explanation:
Let A= | 1 1 1,1 4 5, 1 5 6 | and D= | 7 0 0 ,0 4 0, 0 0 2 | .
Compute AD and DA.
Explain how the columns or rows of A change when A is multiplied by D on the right or on the 1 5 6 0 0 2 left.
Find a 3x3 matrix B, not the identity matrix or zero matrix, such that AB -BA.
Compute AD AD- Compute DA. DA Explain how the columns or rows of A change when A is multiplied by D on the right or on the left.
Choose the correct answer below
A. Right multiplication that is, multiplication on the right by the diagonal matrix D multiplies each row of A by the corresponding diagonal entry of D
B. Both right multiplication that is, multiplication on the right and left multiplication by the diagonal matrix D multiplies each row entry of A by the 0
C. Both right multiplication that is, multiplication on the right and left multiplication by the diagonal matrix D multiplies each column entry of A by the ?
D. Right multiplication that is, multiplication on the right by the diagonal matrix D multiplies each column of A by the corresponding diagonal entry of D Left-multiplication by D multiplies each column of A by the corresponding diagonal entry of D corresponding diagonal entry of D corresponding diagonal entry of D Left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D Find a 3x3 matrix B, not the identity matrix or zero matrix, such that AB BA. Choose the correct answer below There is only one unique solution. B= ?
A. Simplitfy your answers.)
B. There are infinitely many solutions. Any multiple of 13 will satisfy the expression °
C. There does not exist a matrix, B, that will satisfy the expression
Answer:
Check the explanation
Step-by-step explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
After performing the calculations, the right matrix multiplication by D changes each column of A and the left matrix multiplication by D changes each row of A. Also, there are an infinite number of 3x3 matrices that satisfy the condition AB ≠ BA
Explanation:To compute the products AD and DA, we first need to understand how matrix multiplication works. Given matrices A and D as presented, it is important to understand that when A is multiplied by D on the right side (AD), each column of A is multiplied by the corresponding diagonal entry of D. Conversely, when A is multiplied by D on the left side (DA), each row of A is multiplied by the corresponding diagonal entry of D. This is true for any two matrices you want to multiply. Therefore, the correct answer to the first part of the question is statement D.
For the second part of the question, we're seeking a non-identity, non-zero 3x3 matrix B such that AB is not equal to BA. This is generally quite difficult and requires some trial and error. There are an infinite number of matrices B that satisfy the condition AB ≠ BA. The exact form of B depends on A and can be quite complex, but B isn't necessarily unique. Therefore, the correct answer to this question is B: There are infinitely many solutions.
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A company produces fruit juice in 10 different flavors. a local supermarket sells the product, but has only sufficient shelf space to display 3 of the company's 10 fruit juice flavors at a time. How many possible combinations of 3 flavors can the fruit juice company display on the local supermarket shelf?
Answer:
120.
Step-by-step explanation:
This is the number of combinations of 3 from 10.
This = 10C3
= 10! / 3! (10-3)!
= 10*9*8 / 3*2*1
= 120 combinations.
Answer:
120
Step-by-step explanation:
Primary Trigonometric ratios are
used for right angled triangles
only.
True
False
Anything helps!!! :) I need this ASAP!
Answer:
200.96 in^2
Step-by-step explanation:
We want to find the area
A = pi r^2
A = pi (8)^2
A = 3.14 (64)
A =200.96
An analysis of over 160 empirical studies conducted by Norris and her colleagues (2006) revealed evidence of severe to very severe impairment (interference with functioning) among survivors. What percent of survivors were incapacitated?
Answer:
15-25% of survivors were incapacitated
Step-by-step explanation:
According to the analysis conducted by Norris and her colleague in 2006, it was found that out of the 160 empirical studies conducted, 41% showed evidence of severe to very severe impairment (interference with functioning) among disaster survivors.
This in turn corresponds to 15-25% increase in demand for mental health services by the population affected by this disasters.
This percentage of people affected by disasters suffer from disaster's syndrome which include but not limited to shock, bewilderment and void of deep emotion.
Hence they become incapacitated.
The question is incomplete, as the required details are not given. However, I will give a general explanation on how to determine percentages.
To calculate the percentage of survivors that were incapacitated, we need:
The total number of survivors (n)The number of survivors that were incapacitated (k)Assume that:
[tex]n = 250[/tex] --- survivors
[tex]k = 100[/tex] --- survivors that were incapacitated
The percentage (p) is calculated as follows:
[tex]p = \frac kn \times 100\%[/tex]
So, we have:
[tex]p = \frac{100}{250} \times 100\%[/tex]
[tex]p = \frac{100\times 100}{250} \%[/tex]
[tex]p = \frac{10000}{250} \%[/tex]
[tex]p = 40 \%[/tex]
Using the assumed values, 40% of the survivors were incapacitated.
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One study on managers’ satisfaction with management tools reveals that 58% of all managers use self-directed work teams as a management tool. Suppose 70 managers selected randomly in the United States are interviewed. What is the probability that fewer than 35 use self-directed work teams as a management tool?
Answer:
6.94% probability that fewer than 35 use self-directed work teams as a management tool
Step-by-step explanation:
I am going to use the normal approximation to the binomial to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/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.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 70, p = 0.58[/tex]
So
[tex]\mu = E(X) = np = 70*0.58 = 40.6[/tex]
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{70*0.58*0.42} = 4.13[/tex]
What is the probability that fewer than 35 use self-directed work teams as a management tool?
Using continuity correction, this is P(X < 35 - 0.5) = P(X < 34.5), which is the pvalue of Z when X = 34.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{34.5 - 40.6}{4.13}[/tex]
[tex]Z = -1.48[/tex]
[tex]Z = -1.48[/tex] has a pvalue of 0.0694.
6.94% probability that fewer than 35 use self-directed work teams as a management tool
The question requires calculation of binomial probability. Given that the rate of success is 58% (or 0.58) and we're trying to find the likelihood of fewer than 35 successes out of 70 trials, one must sum the binomial probabilities from k=0 to 34.
Explanation:This problem is about calculation of binomial probability, which is a specific type of probability that deals with experiments that have two possible outcomes: success (in this case, using self-directed work teams) or failure (not using self-directed work teams). Given that the rate of success is 58% (or 0.58 as a decimal), and we're looking for the probability of fewer than 35 successes out of 70 trials (or managers), we can solve using the formula for binomial probability.
The basic form of the binomial formula is: [tex]P(X=k) = C(n, k) * (p^k) * ((1-p)^{(n-k)})[/tex], where n is the number of trials, k is the number of successful trials, p is the probability of success, and C(n, k) is a combination which calculates the number of possible combinations of n items taken k at a time.
To find P(X<35), we sum from k=0 to 34. Thus, this involves a fair amount of calculation, and you may want to use software or a calculator that has binomial probability functionality.
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What is the area of the wall that will be painted
Answer:
B. 104
Step-by-step explanation:
Just find the area of the wall and subtract the area of the window. The area of the wall is 10 times 11 which is 110. The area of the window is 2 times 3 which is 6. 110 minus 6 is 104.
f(x) = x4 - 50x2 + 3 (a) Find the intervals on which f is increasing. (Enter the interval that contains smaller numbers first.) ( , ) ∪ ( , ) Find the intervals on which f is decreasing. (Enter the interval that contains smaller numbers first.) ( , ) ∪ ( , ) (b) Find the local minimum and maximum values of f. (min) (max) (c) Find the inflection points. ( , ) (smaller x value) ( , ) (larger x value) Find the intervals on which f is concave up. (Enter the interval that contains smaller numbers first.) ( , ) ∪ ( , ) Find the interval on which f is concave down. ( ,
Answer:
(-5, 0) ∪ (5, ∞)
Step-by-step explanation:
I find a graph convenient for this purpose. (See below)
__
When you want to find where a function is increasing or decreasing, you want to look at the sign of the derivative. Here, the derivative is ...
f'(x) = 4x^3 -100x = 4x(x^2 -25) = 4x(x +5)(x -5)
This has zeros at x=-5, x=0, and x=5. The sign of the derivative will be positive when 0 or 2 factors have negative signs. The signs change at the zeros. So, the intervals of f' having a positive sign are (-5, 0) and (5, ∞).
The details provided do not correspond to a coherent mathematics problem regarding the function [tex]f(x) = x^4 - 50x^2 + 3[/tex]. Therefore, an accurate response cannot be provided without further information or correct context.
Explanation:To address the question about the function [tex]f(x) = x^4 - 50x^2 + 3[/tex], we need to analyze its intervals of increase and decrease, as well as find any local extrema and points of concavity. However, the question as posed does not provide enough context or coherent detail for the actions requested, such as finding the inflection points or intervals of concavity, since no specific function was clearly defined. Instead, various unrelated Mathematics problems are listed, each of which is missing comprehensive details needed to provide an accurate answer.
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The head of institutional research at a university believed that the mean age of full-time students was declining. In 1995, the mean age of a full-time student was known to be 27.4 years. After looking at the enrollment records of all 4934 full-time students in the current semester, he found that the mean age was 27.1 years, with a standard deviation of 7.3 years. He conducted a hypothesis of Upper H 0: muequals27.4 years versus Upper H 1: muless than27.4 years and obtained a P-value of 0.0020. He concluded that the mean age of full-time students did decline. Is there anything wrong with his research?
Answer:
[tex]t=\frac{27.1-27.4}{\frac{7.3}{\sqrt{4934}}}=-2.887[/tex]
The degrees of freedom are given by:
[tex] df= n-1 = 4934-1= 4933[/tex]
Then the p value for this case calculated as:
[tex]p_v =P(t_{4933}<-2.887) =0.002[/tex]
Since the p value is a very lower value using any significance level for example 1% or 5% we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significanctly less than 27.4. So then is not anything wrong with the conclusion
Step-by-step explanation:
Information provided
[tex]\bar X=27.1[/tex] represent the sample mean
[tex]s=7.3[/tex] represent the sample standard deviation
[tex]n=4934[/tex] sample size
[tex]\mu_o =27.4[/tex] represent the value to test
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to verify if the mean age of full-time students did decline (less than 27.4), the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 27.4[/tex]
Alternative hypothesis:[tex]\mu < 27.4[/tex]
The statistic for this case is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info we got:
[tex]t=\frac{27.1-27.4}{\frac{7.3}{\sqrt{4934}}}=-2.887[/tex]
The degrees of freedom are given by:
[tex] df= n-1 = 4934-1= 4933[/tex]
Then the p value for this case calculated as:
[tex]p_v =P(t_{4933}<-2.887) =0.002[/tex]
Since the p value is a very lower value using any significance level for example 1% or 5% we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significanctly less than 27.4. So then is not anything wrong with the conclusion
Find the solution for system of equations
2x-3y=2 x=6y-5
Harry is trying to solve the equation y = 2x2 − x − 6 using the quadratic formula. He has made an error in one of the steps below. Find the step where Harry went wrong. (1 point)
Step 1: x equals the negative of negative 1 plus or minus the square root of the quantity negative one squared minus 4 times 2 times negative six, end quantity, all over 2 times 2.
Step 2: x equals the negative of negative 1 plus or minus the square root of negative one plus forty-eight all over two times 2.
Step 3: x equals the negative of negative 1 plus or minus the square root of forty-seven all over two times 2.
Step 4: x equals 1 plus or minus the square root of forty-seven all over 4.
Answer:
Step 2
Step-by-step explanation:
The quadratic formula is given by [tex]x=-b+-\frac{\sqrt{b^{2}-4ac } }{2a}[/tex]
Our equation is y = 2x²-x-6
So here our a = 2, b = -1, and c = -6
We can now plug these numbers into our formula
[tex]x= -(-1) +-\frac{\sqrt{(-1)^{2}-4(2)(-6) } }{2(2)} = 1 +-\frac{\sqrt{1+24} }{4} = 1+-\frac{\sqrt{25} }{4}[/tex]
Step 2 is incorrect because it states that "x equals the negative of negative 1 plus or minus the square root of negative one plus forty-eight all over two times 2."
The correct statement would be "x equals the negative of negative 1 plus or minus the square root of positive one plus forty-eight all over two times 2.", because the square of a negative is positive, resulting in a positive one.
Since this step is incorrect, the steps after are also incorrect, but Harry went wrong at Step 2
Answer:
The above answer is correct.
Step-by-step explanation:
I got it right on the test
The combined math and verbal scores for students taking a national standardized examination for college admission, is normally distributed with a mean of 820 and a standard deviation of 200. If a college requires a student to be in the top 15 % of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission at the college calculator
Answer: 1007.28
Step-by-step explanation:
Given : The combined math and verbal scores for students taking a national standardized examination for college admission, is normally distributed with
[tex]\mu=820\ \ \ ,\ \sigma=200[/tex]
If a college requires a student to be in the top 15 % of students taking this test, it means that they want the students that score 85 percentile or above.
Let X be the scores of any random student, we require
[tex]P(X<x)=0.85[/tex], where x is minimum score that such a student can obtain and still qualify for admission at the college.
Formula for z-score = [tex]z=\frac{x-\mu}{\sigma}=\frac{x-820}{200}[/tex] ...(i)
From normal z-value table , [tex]P(z<1.036)=0.85[/tex]...(ii)
From (i) and (ii) , we get
[tex]\frac{x-820}{200}=1.0364\\\\\Rightarrow\ x-820=207.28\\\\\Rightarrow\ x=207.28+820=1007.28[/tex]
Hence, the minimum score that such a student can obtain and still qualify for admission at the college is 1007.28.
You have decided to stop drinking Starbucks coffee and invest that money in an IRA. If you
deposit $
Complete question :
You have decided to stop drinking Starbucks coffee and invest that money in an IRA. If you deposit $492 each month earning 6% interest, how much will you have in the account after 40 years?
Answer:
$250,329.60
Step-by-step explanation:
Given:
Principal, P= $492 per month
= $492 * 12 = $5,904 per year
Rate, R= 6%
Time, T= 40 years
Let's first find the amount after 40 years.
Amount = 5940 * 40 = $236,160
The interest after 40 years =
[tex] Interest = \frac{PRT}{100} = \frac{5904 * 6 * 40}{100}[/tex]
Interest = $14,169.60
The total amount I will have in my account after 40 years:
Amount + interest
= $236,160 + $14,169.60
= $250,329.60
The following are quality control data for a manufacturing process at Kensport Chemical Company. The data show the temperature in degrees centigrade at five points in time during a manufacturing cycle. Sample x R 1 95.72 1.0 2 95.24 0.9 3 95.18 0.9 4 95.44 0.4 5 95.46 0.5 6 95.32 1.1 7 95.40 0.9 8 95.44 0.3 9 95.08 0.2 10 95.50 0.6 11 95.80 0.6 12 95.22 0.2 13 95.56 1.3 14 95.22 0.6 15 95.04 0.8 16 95.72 1.1 17 94.82 0.6 18 95.46 0.5 19 95.60 0.4 20 95.74 0.6 The company is interested in using control charts to monitor the temperature of its manufacturing process. Compute the upper and lower control limits for the R chart. (Round your answers to three decimal places.) UCL
Answer: The upper control limit is 1.40581
The lower control limit is 0.000
Explanation: Check attachment
The upper control limit (UCL) for the R chart is 1.986 and the lower control limit (LCL) is 0.
To compute the upper and lower control limits for the R chart, we need to calculate the average range (R-bar) and the control limits.
First, let's calculate the range (R) for each sample by subtracting the smallest value from the largest value:
Sample 1: 1.0
Sample 2: 0.9
Sample 3: 0.9
Sample 4: 0.4
Sample 5: 0.5
Sample 6: 1.1
Sample 7: 0.9
Sample 8: 0.3
Sample 9: 0.2
Sample 10: 0.6
Sample 11: 0.6
Sample 12: 0.2
Sample 13: 1.3
Sample 14: 0.6
Sample 15: 0.8
Sample 16: 1.1
Sample 17: 0.6
Sample 18: 0.5
Sample 19: 0.4
Sample 20: 0.6
Next, calculate the average range (R-bar) by summing up all the ranges and dividing by the number of samples:
R-bar = (1.0 + 0.9 + 0.9 + 0.4 + 0.5 + 1.1 + 0.9 + 0.3 + 0.2 + 0.6 + 0.6 + 0.2 + 1.3 + 0.6 + 0.8 + 1.1 + 0.6 + 0.5 + 0.4 + 0.6) / 20
R-bar = 0.735
To calculate the upper control limit (UCL) for the R chart, multiply the R-bar by a constant factor. For small sample sizes like this (n=5), the constant factor is typically 2.704:
UCL = R-bar * 2.704
UCL = 0.735 * 2.704
UCL = 1.986
Finally, to calculate the lower control limit (LCL) for the R chart, multiply the R-bar by another constant factor. For small sample sizes like this, the constant factor is typically 0:
LCL = R-bar * 0
LCL = 0
Therefore, the upper control limit (UCL) for the R chart is 1.986 and the lower control limit (LCL) is 0.
To Learn more about constant factor here:
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A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The p-value is a. .9772. b. .5475. c. 2.000. d. .0228.
Answer:
Step-by-step explanation:
Given data
Average sales = 8000
n = 64
standard deviation = 1200
8300
The solution is attached in the picture below
a scientist counted 11 crows to every 3 hawks. if this data holds true, how many hawks would he expect to see if there were 363 crows?
Answer:
99
Step-by-step explanation:
11 crows : 3 hawks
363 crows: X hawks
X/363 = 3/11
X = 363 × 3/11
X = 99
The A.C. Nielsen Company collected data on the weekly TV viewing times, in hours, of 200 people. Suppose that the sample mean is 30.25, the sample standard deviation is 12.60, and that the histogram of the viewing times is bell shaped. Approximately what percent of the people in the study will have weekly TV viewing times between 17.65 and 42.85
Answer:
By the Empirical Rule, approximately 68% the people in the study will have weekly TV viewing times between 17.65 and 42.85
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed(bell-shaped) random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 30.25
Standard deviation = 12.60
Approximately what percent of the people in the study will have weekly TV viewing times between 17.65 and 42.85
17.65 = 30.25 - 1*12.60
So 17.65 is one standard deviation below the mean.
42.85 = 30.25 + 1*12.60
So 42.85 is one standard deviation above the mean
By the Empirical Rule, approximately 68% the people in the study will have weekly TV viewing times between 17.65 and 42.85
What is the equation of the following graph?
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