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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Vermont-based Green Mountain Coffee Roasters dominates the market for single-serve coffee in the United States, with its subsidiary Keurig accounting for approximately 70% of sales ("Rivals Try to Loosen Keurig's Grip on Single-Serve Coffee Market," Chicago Tribune, February 26, 2011). But Keurig's patent on K-cups, the plastic pods used to brew the coffee, is expected to expire in 2012, allowing other companies to better compete. Suppose a potential competitor has been conducting blind taste tests on its blend and finds that 47% of consumers strongly prefer its French Roast to that of Green Mountain Coffee Roasters. After tweaking its recipe, the competitor conducts a test with 144 tasters, of which 72 prefer its blend. The competitor claims that its new blend is preferred by more than 47% of consumers to Green Mountain Coffee Roasters' French Roast.
Refer to Exhibit 9-7. At the 1% significance level, does the evidence support the claim?
a. No, since the value of the test statistic is less than the critical value
b. Yes, since the value of the test statistic is less than the critical value
c. No, since the value of the test statistic is greater than the critical value
d. Yes, since the value of the test statistic is greater than the critical value
Answer:
a. No, since the value of the test statistic is less than the critical value
Step-by-step explanation:
1) Data given and notation
n=144 represent the random sample taken
X=72 represent the number of people that prefer the blend
[tex]\hat p=\frac{72}{144}=0.5[/tex] estimated proportion of people that prefer the blend
[tex]p_o=0.47[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.959
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion if higher than 0.47:
Null hypothesis:[tex]p\leq 0.47[/tex]
Alternative hypothesis:[tex]p > 0.47[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.5 -0.47}{\sqrt{\frac{0.47(1-0.47)}{144}}}=0.721[/tex]
4) Statistical decision
We can calculate the critical value since we have a right tailed test, we need to look into the normal standard distribution a value that accumulates 0.01 of the area on the right and 0.99 on the left. And this value is:
[tex]z_{\alpha/2}=2.33[/tex]
And we can use the following excel code to find the critical value: "=NORM.INV(0.99,0,1)"
Our calculated value on this case is less than the critical value so the best conclusion is:
a. No, since the value of the test statistic is less than the critical value
Why are line extensions more common than new products?a) They are higher risk and more expensive.b) They are guaranteed to succeed in the marketplace.c) They are less expensive and lower-risk.d) They require no market research.e) Modifications to existing products are very easy
Answer:
The answer is c): They are less expensive and lower-risk.
Step-by-step explanation:
Line extensions are more common than new products because they do not require an overhaul of old or existing products or marketing strategies: old/existing products only have to be changed, and this requires lesser money, and is less risky. On the other hand, creating new products is more expensive and risky than line extensions.
A farmer uses a lot of fertilizer to grow his crops. The farmer’s manager thinks fertilizer products from distributor A contain more of the nitrogen that his plants need than distributor B’s fertilizer does. He takes two independent samples of four batches of fertilizer from each distributor and measures the amount of nitrogen in each batch. Fertilizer from distributor A contained 23 pounds per batch and fertilizer from distributor B contained 18 pounds per batch. Suppose the population standard deviation for distributor A and distributor B is four pounds per batch and five pounds per batch, respectively. Assume the distribution of nitrogen in fertilizer is normally distributed. Let µ1and µ2 represent the average amount of nitrogen per batch for fertilizer’s A and B, respectively. Which of the following is the appropriate conclusion at the 5% significance level? The test statistic calculated in Excel with these data is 1.5617.
Answer:
[tex]z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617[/tex]
[tex]p_v =P(Z>1.5617)=0.059[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X_{A}=23[/tex] represent the mean for the sample A
[tex]\bar X_{B}=18[/tex] represent the mean for the sample B
[tex]\sigma_{A}=4[/tex] represent the population standard deviation for the sample A
[tex]\sigma_{B}=5[/tex] represent the population standard deviation for the sample B
[tex]n_{A}=4[/tex] sample size selected A
[tex]n_{B}=4[/tex] sample size selected B
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean for A is higher than the mean for B, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{A}-\mu_{B}\leq 0[/tex]
Alternative hypothesis:[tex]\mu_{A}-\mu_{B}>0[/tex]
We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:
[tex]z=\frac{(\bar X_{A}-\bar X_{B})-0}{\sqrt{\frac{\sigma^2_{A}}{n_{A}}+\frac{\sigma^2_{B}}{n_{B}}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617[/tex]
P-value
Since is a one right tailed test the p value would be:
[tex]p_v =P(Z>1.5617)=0.059[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.
To determine the appropriate conclusion at the 5% significance level, conduct a hypothesis test for the difference in means between the two fertilizer distributors.
Explanation:To determine the appropriate conclusion at the 5% significance level, we need to conduct a hypothesis test for the difference in means between the two fertilizer distributors. The test statistic calculated in Excel is 1.5617. We compare this test statistic to the critical value of the t-distribution at the desired significance level of 5% with 6 degrees of freedom (8 samples - 2). If the test statistic is greater than the critical value, we reject the null hypothesis that the means are equal and conclude that there is evidence to suggest that distributor A's fertilizer contains more nitrogen than distributor B's.
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Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean resistance. Find the standard deviation of the resistances. Find the probability that the resistance is between 98 and 102 Ω. Suppose that resistances of different resistors are independent. What is the probability that three out of six resistors have resistances greater than 100 Ω?
Answer:
[tex]E[R][/tex] = 99 Ω
[tex]\sigma_R[/tex] = 2.3094 Ω
P(98<R<102) = 0.5696
Step-by-step explanation:
The mean resistance is the average of edge values of interval.
Hence,
The mean resistance, [tex]E[R] = \frac{a+b}{2} = \frac{95+103}{2} = \frac{198}{2}[/tex] = 99 Ω
To find the standard deviation of resistance, we need to find variance first.
[tex]V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333[/tex]
Hence,
The standard deviation of resistance, [tex]\sigma_R = \sqrt{V(R)} = \sqrt5.333[/tex] = 2.3094 Ω
To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.
[tex]z_1 = \frac{102-99}{2.3094} = 1.299[/tex]
[tex]z_2 = \frac{98-99}{2.3094} = -0.433[/tex]
From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696
which of the following number sets does 25 belong in?
2 and 4
all of the above
3 and 5
1 and 2
Answer:
all of the above
Step-by-step explanation:
The number 25 is a natural number as it belongs to the set [1,2,3,4,5,......]
The number 25 is a whole number as it belongs to the set [0,1,2,3,4,5,......]
The number 25 is an Integer as it belongs to the set [...,-5,-4,-3,-2,-1,0,1,2,3,4,5,...]
The number 25 is a rational number as it can be expressed as [tex]\[\frac{25}{1}\][/tex]
For the same reason , number 25 is a real number as it belongs to the set of rational numbers.
So the correct option is "all of the above".
25 does not belong to any of the given number sets (2 and 4, 1 and 2, or 3 and 5).
Explanation:
The number 25 does not belong to any of the provided number sets i.e. 2 and 4, 1 and 2 or 3 and 5. A number set typically refers to a collection of numbers, and in this case, 25 is absent in all the provided sets. The given number sets only contain the numbers 1, 2, 3, 4 and 5. Thus, 25 does not belong to any of these sets.
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In a random sample of 9 residents of the state of Florida, the mean waste recycled per person per day was 2.4 pounds with a standard deviation of 0.75 pounds. Determine the 80% confidence interval for the mean waste recycled per person per day for the population of Florida. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
When the sample size is small (< 30) and the population standard deviation is unknown , then we use t-test.
The confidence interval for population mean will be :
[tex]\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}[/tex] (1)
, where [tex]\overline{x}[/tex] = sample mean
t* = Critical value (based on degree of freedom and significance level).
s= sample standard deviation
n= sample size.
As per given we have
n= 9
Degree of freedom = n-1 = 8
[tex]\overline{x}=2.4[/tex]
s= 0.75
Significance level =[tex]\alpha=1-0.80=0.20[/tex]
Using students' t distribution table ,
Critical value : [tex]t^*=t_{\alpha/2,df}=t_{0.10,8}=1.3304[/tex]
We assume that the population is approximately normal.
Then, a 80% confidence interval for the mean waste recycled per person per day for the population of Florida will be :
[tex]2.4\pm (1.3304)\dfrac{0.75}{\sqrt{8}}[/tex] (Substitute the values in (1))
[tex]2.4\pm (1.3304)\dfrac{0.75}{2.82842712475}[/tex]
[tex]2.4\pm (1.3304)(0.265165042945)[/tex]
[tex]2.4\pm 0.352775573134\approx2.4\pm0.353=(2.4-0.353,\ 2.4+0.353)=(2.047,\ 2.753)[/tex]
Hence, the 80% confidence interval for the mean waste recycled per person per day for the population of Florida. = (2.047, 2.753)
a community program choose 16 fifth grade students every year and provide each of them with the same amount of money to attend music or art camp last year the program awarded a total of 8,400 to the students how much
Answer: 525
Step-by-step explanation: As I read the question I’m getting the idea of division. The community program chooses 16 students every year. The 8,400 dollars from last year was the amount of money the students receive all together. Therefore 8,400 divided by 16 is 525
Final answer:
Each of the 16 fifth grade students received $525 from the community program to attend music or art camp, calculated by dividing the total funds of $8,400 by 16 students.
Explanation:
The question asked is about calculating the amount of money awarded to each of the 16 fifth grade students by a community program for attending music or art camp. Since the program awarded a total of $8,400 last year and 16 students were chosen, we need to perform a simple division to find out how much money each student received. To do this, we divide the total amount of money ($8,400) by the number of students (16).
Step-by-step Calculation:
Divide the total amount of money by the number of students: $8,400 \/ 16.
Calculate the result to determine the amount per student.
Therefore, each student received $525 to attend the music or art camp.
In a group of mherchants, 80% of them purchase goods from Asia, and 25% of them purchase goods from Europe. Which of following statement is individually sufficient to calculate what percent of the merchants in the group purchase goods from Europe but not form Asia? 7. 25% of the merchants who purchase goods from Asia also purchase from Europe. 15% of all merchants purchase goods from neither Asia nor Europe 0% of all merchants purchase good from both Asia and Europe
Answer:
7. 25% of the merchants who purchase goods from Asia also purchase from Europe.
Step-by-step explanation:
I am going to say that:
A is the percentage of merchants who purchase goods from Asia.
B is the percentage of merchants who purchase goods from Europe.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a merchant purchases goods from Asia but not from Europe and [tex]A \cap B[/tex] is the probability that a merchant purchases goods from both Asia and Europe.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
Which of following statement is individually sufficient to calculate what percent of the merchants in the group purchase goods from Europe but not form Asia?
We already have B.
Knowing [tex]A \cap B[/tex], that is, the percentage of those who purchase from both Asia and Europe, we can find b.
So the correct answer is:
7. 25% of the merchants who purchase goods from Asia also purchase from Europe.
For results based on a small random sample from a bell-shaped distribution, the distribution of the sample mean is
A. approximately a normal distribution.
B. not a bell-shaped distribution.
C. a uniform distribution.
D. approximately a standard normal (z-score) distribution
Answer:
A. approximately a normal distribution.
Step-by-step explanation:
There may be a few differences, but the sampling distribution of the sample mean is still approximately normal.
So the correct answer is:
A. approximately a normal distribution.
Answer:
Correct answer is (A) {Normal distribution}
Step-by-step explanation:
sampling distribution of the sample mean is still approximately normal.
Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.
This equation is separable, as
[tex]\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx[/tex]
Integrate both sides; on the left, expand the fraction as
[tex]\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)[/tex]
Then
[tex]\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C[/tex]
[tex]\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C[/tex]
Since [tex]y(0)=1[/tex], we get
[tex]\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1[/tex]
so that the particular solution is
[tex]\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}[/tex]
A researcher is interested in developing a model that can be used to distribute assistance to low-income families for food costs. She used data from a national social survey to predict weekly amount spent on food using household income (in $1000). The resulting regression equation is ModifyingAbove Food divided by wk with caret equals 101.33 plus 0.77 HIncome.Food/wk=101.33+0.77HIncome. How much money would be needed to feed a family for a week whose household income is $12,000?
The estimated cost to feed a family for a week with a household income of $12,000 would be $9,341.33.
Explanation:To find out how much money would be needed to feed a family for a week whose household income is $12,000, we need to use the regression equation provided. The equation is Food/wk = 101.33 + 0.77HIncome. We substitute the value of HIncome with $12,000 and solve for Food/wk.
Food/wk = 101.33 + 0.77(12,000)
Food/wk = 101.33 + 9240
Food/wk = $9,341.33
Therefore, the estimated cost to feed a family for a week with a household income of $12,000 would be $9,341.33.
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You can buy a television for $349 cash or pay $75 down and the balance in 18 monthly payments of 22.50 is the installment price of the TV? By what percent would the installment price be greater than the cash price?
Answer:
Step-by-step explanation:
If you pay cash, the total amount that you will pay for the television is $349
If you pay $75 down, the balance would be paid in 18 monthly payments of 22.50 which is the installment price of the TV. Total amount paid in 18 months would be
22.5 × 18 = $405
Total cost of the TV when you pay in installments would be
405 + 75 = $480
Difference between the installment price and the cash price would be
480 - 349 = $131
The percent by which the installment price would be greater than the cash price is
131/349 × 100 = 37.5%
According to a research paper, the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in a particular city is approximately normally distributed with a mean of 18.4 ppm. Suppose that the standard deviation of carbon monoxide exposure is 5.9 ppm.
The probability of a motorbike rider in this city experiencing carbon monoxide exposure exceeding 20 ppm is 0.3932, while the probability of exceeding 25 ppm is 0.1316, based on the normal distribution with a mean of 18.6 ppm and a standard deviation of 5.9 ppm.
Define the random variable X as the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in this city.
Since the distribution of X is normally distributed with a mean of 18.6 ppm and a standard deviation of 5.9 ppm, we can use the standard normal distribution to calculate probabilities.
To find the probability of X being greater than 20 ppm, we need to find the area to the right of 20 ppm under the standard normal curve.
We can calculate this area using a z-score, which is defined as the number of standard deviations a specific point is away from the mean. In this case, the z-score for 20 ppm is (20 ppm - 18.6 ppm) / 5.9 ppm = 0.271.
Using a standard normal table or calculator, we can find that the area to the right of 0.271 is 0.3932.
Therefore, the probability of someone riding a motorbike for 5 km on a highway in this city experiencing a carbon monoxide exposure of more than 20 ppm is 0.3932.
Follow the same steps as in part a, but use a z-score of (25 ppm - 18.6 ppm) / 5.9 ppm = 1.119.
The area to the right of 1.119 under the standard normal curve is 0.1316.
Therefore, the probability of someone riding a motorbike for 5 km on a highway in this city experiencing a carbon monoxide exposure of more than 25 ppm is 0.1316.
Complete question:
According to a research paper, the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in a particular city is approximately normally distributed with a mean of 18.6 ppm. Suppose that the standard deviation of carbon monoxide exposure is 5.9 ppm (a) Approximately what proportion of those who ride a motorbike for 5 km on a highway in this city will experience a carbon monexide exposure of more than 20 ppm? (Round your answer to fou decimal places.) (b) Approximately what proportion of those who ride a motorbike for 5 km on a highway in this city will experience a carbon monoxide exposure of more than 25 ppm?
A tank contains 6,000 L of brine with 16 kg of dissolved salt. Pure water enters the tank at a rate of 60 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. (a) How much salt is in the tank after t minutes? y = kg (b) How much salt is in the tank after 20 minutes? (Round your answer to one decimal place.) y = kg
Answer:
a) [tex]S_{a}(t)=16Kg-0.16Kg*\frac{t}{min}[/tex]
b) [tex]S_{a}(20)=12.8Kg[/tex]
Step-by-step explanation:
It can be seen in the graph that the water velocity and solution velocity is the same, but the salt concentation will be lower
Water velocity [tex]V_{w} = 60\frac{L}{min}[/tex]
Solution velocity [tex]V_{s} = 60\frac{L}{min}[/tex]
Brine concentration = [tex]\frac{6,000L}{16Kg}=375\frac{L}{Kg}[/tex]
a) Amount of salt as a funtion of time Sa(t)
[tex]S_{a}(t)=16Kg-\frac{60Kg*L}{375L}*\frac{(t)}{min}=[tex]16Kg-0.16Kg*\frac{t}{min}[/tex]
b) [tex]S_{a}(20)=16Kg-0.16\frac{Kg}{min}*(20min)=16Kg-3.2Kg=12.8Kg[/tex]
This value was to be expected since as the time passes the concentration will be lower due to the entrance to the pure water tank
To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min and then calculate the amount of salt in the tank after a specific time.
Explanation:To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min:
Rate of salt entering = (60 L/min) * (16 kg/6000 L) = 0.16 kg/min
Therefore, the amount of salt in the tank after t minutes is given by:
y = 0.16 kg/min * t min = 0.16t kg
For a school field trip the students had two options for lunch, a turkey or egg salad sandwich, so it is impossible for a student have both lunches. If the probability that a student chooses a turkey sandwich is 0.10, and the probability that a student chooses an egg salad sandwich is 0.67, what is the probability that a student chooses a turkey or egg salad sanwich?
Answer: 0.77
Step-by-step explanation:
Given : Probability that a student chooses a turkey sandwich is
P(Turkey )= 0.10
Probability that a student chooses an egg salad sandwich is
P(egg salad)=0.67
Also, it is impossible for a student have both lunches.
∴ P(Turkey and egg salad) =0
Now , the probability that a student chooses a turkey or egg salad sandwich will be
P(Turkey or egg salad) = P(Turkey )+ P(egg salad)- P(Turkey and egg salad)
= 0.10+ 0.67-0 = 0.77
Hence, the probability that a student chooses a turkey or egg salad sandwich= 0.77
Final answer:
The probability that a student picks either a turkey or an egg salad sandwich for their school field trip is 0.77 or 77%.
Explanation:
To calculate the probability that a student chooses either a turkey or egg salad sandwich for lunch, we use the formula for the probability of an 'or' event.
Since the options are mutually exclusive, meaning a student can only choose one type of sandwich, we simply add the individual probabilities together.
The probability of choosing a turkey sandwich is 0.10 and the probability of choosing an egg salad sandwich is 0.67.
Therefore, we can calculate it as follows:
P(turkey OR egg salad) = P(turkey) + P(egg salad)
P(turkey OR egg salad) = 0.10 + 0.67
P(turkey OR egg salad) = 0.77
So the probability that a student picks either a turkey or an egg salad sandwich is 0.77, or 77%.
Lin and Priya were working on solving this system of equations.
{1/3x+2y=4
x+y=-3
Lin's first move is to multiply the first equation by 3.
Priya's first move is to multiply the second equation by 2.
Explain why either move creates a new equation with the same solutions as the original equation.
The first equation is x + 6y = 36
the second equation is x+y = -6
(x,y) =(-6,3)
Answer:
we can conclude two things that:
If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.Either person's move will work. Lin's move eliminated the x variable, while Priya's eliminated y variable, but in the end the solution was same.Step-by-step explanation:
Why either move creates a new equation with the same solutions as the original equation?
If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.
When we multiple the two sides of any given equation by the same number, it would keep the two sides of that particular equation equal. So, whatever the the solution the first equation may get, will still work for the second equation.
Determining Lin's first move i.e. to multiply the first equation by 3.
Let us consider the equation
x/3 + 2y = 4 .....[1]
x + y = -3 .....[2]
Lin's first move is to multiply the first equation by 3.
3(x/3 + 2y) = 3(4 )
x + 6y = 12 .....[3]
Now subtract the Equation [2] from Equation [3]
x + 6y - x - y = 12 - (-3)
5y = 15
y = 3
Putting y = 3 in [2]
x + (3) = -3
x = -6
So, x = -6 and y = 3
Determining Priya's first move i.e. to multiply the Second equation by 2.
Let us consider the equation
x/3 + 2y = 4 .....[1]
x + y = -3 .....[2]
Priya's first move is to multiply the second equation by 2.
2(x + y) = 2(-3)
2x + 2y = -6 .....[3]
Now subtract the Equation [2] from [1]
x/3 + 2y - 2x - 2y= 4 - (-6)
x/3 - 2x = 10
x - 6x = 30
x = -6
Putting x = -6 in Equation [2]
x + y = -3
-6 + y = -3
y = 3
So, x = -6 and y = 3
So, from the entire analysis, we can conclude two things that:
If we multiple the two sides of any given equation by the same factor, we would get an equivalent equation, which will have the same solution as the original solution.Either person's move will work. Lin's move eliminated the x variable, while Priya's eliminated y variable, but in the end the solution was same.Keywords: system of equation, solution, equation
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A political scientist wants to know how college students feel about the social security system. She obtains a list of the 3114 undergraduates at her college and mails a questionnaire to 250 students selected at random. Only 100 of the questionnaires are returned. In this study, the rate of non-response would be a. 0.25. b. 0.40. x. 0.75. d. 0.60
Answer: d. 0.60
Step-by-step explanation:
When are performing sample surveys , when the selected participant is giving any response is denoted as non - response.
The proportion of these participants of the sample is known as the non-response rate.
Given : A political scientist wants to know how college students feel about the social security system.
She obtains a list of the 3114 undergraduates at her college and mails a questionnaire to 250 students selected at random.
i.e. Sample size : n= 290
Only 100 of the questionnaires are returned.
Individual gave response =100
Individual gave no-response =250-100 =150
The rate of non-response [tex]=\dfrac{\text{Individual gave no-response}}{n}[/tex]
[tex]=\dfrac{150}{250} =0.60[/tex]
Hence, the rate of non-response would be 0.60 .
Thus , the correct option is d. 0.60.
what transformations are represented by the following coordinate graphing? (geometry)
(a,b) --> (a,-b)
(a,b) --> (a, b+5)
(a,b) --> (b,-a)
Answer:
(a,b) → (a,-b) : Reflection about x axis.
(a,b) → (a, b+5) : Translation of the point by 5 units up.
(a,b) → (b,-a) : Rotation by 90 degree clockwise.
Step-by-step explanation:
Given:
The transformation of points are given as:
(a,b) → (a,-b)
(a,b) → (a, b+5)
(a,b) → (b,-a)
Now, let us consider each transformation one by one.
(1) (a,b) → (a,-b)
Here, the order of the coordinates has not changed. But, the y coordinate of the point has changed. The y coordinate was 'b' and it has changed only its sign but not value. So, it is a transformation related to reflection.
In reflection, only the sign changes. Since, the 'y' coordinate sing is reversed, so, it is a reflection about x axis.
(2) (a,b) → (a, b+5)
Here, the 'y' coordinate of the point has changed. The change is from 'b' to 'b+5'. So, 5 is added to the y coordinate. As per transformation rules, if a positive number 'C' is added to the y coordinate, then the point shifts vertically up by 'C' units. Hence, there is a translation of 5 units up here.
(3) (a,b) → (b,-a)
Here, the 'x' and 'y' coordinates interchange their values and also the new y coordinate has its sign reversed. This happens in rotation.
We know that, (x, y) → (y, –x) is true when there is rotation by 90 degree clockwise.
So, the point (a,b) → (b,-a) is rotated by 90 degree clockwise.
Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (σ1= 2 ), so that the standard error of the difference between means is obtained by pooling the sample variances. A paint manufacturer wanted to compare the drying times of two different types of paint. Independent simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows.Type A: X1= 71.5hr, S1=3.4 hr N1=11Type B: X2=68.5 hr, S2= 3.6 hr, N2= 9Construct a 99% confidence interval for μ1-μ2 , the difference between the mean drying time for paint type A and the mean drying time for paint type B.
Answer:
The indicated confidence interval for the difference between the two population means is (-1.5159, 7.5159)
Step-by-step explanation:
Let the drying times of type A be the first population and the drying times of type B be the second population. Then
We have small sample sizes [tex]n_{1} = 11[/tex] and [tex]n_{2} = 9[/tex], besides [tex]\bar{x}_{1} = 71.5[/tex], [tex]s_{1} = 3.4[/tex] , [tex]\bar{x}_{2} = 68.5[/tex] and [tex]s_{2} = 3.6[/tex]. Therefore, the pooled
estimate is given by
[tex]s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(11-1)(3.4)^{2}+(9-1)(3.6)^{2}}{11+9-2} = 12.1822[/tex]
The 99% confidence interval for the true mean difference between the mean drying time of type A and the mean drying time of type B is given by
[tex](\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{11}+\frac{1}{9}}[/tex], i.e.,
[tex](71.5-68.5)\pm t_{0.005}(3.4903)\sqrt{\frac{1}{11}+\frac{1}{9}}[/tex]
where [tex]t_{0.005}[/tex] is the 0.5th quantile of the t distribution with (11+9-2) = 18 degrees of freedom. So
[tex]3\pm(-2.8784)(3.4903)(0.4495)[/tex], i.e.,
the indicated confidence interval for the difference between the two population means is (-1.5159, 7.5159)
IQ scores (as measured by the Stanford-Binet intelligence test) are normally distributed with a mean of 100 and a standard deviation of 19.Find the approximate number of people in the United States (assuming a total population of 280,000,000) with an IQ higher than 140. (Round your answer to the nearest hundred thousand.)
Answer:
[tex]P(X>140)=1-P(X\leq 140)=1-P(\frac{X-\mu}{\sigma}<\frac{140-\mu}{\sigma})=1-P(Z<\frac{140-100}{19})=1-P(Z<2.105)=0.0176[/tex]
Number =0.0176*280000000=4928000 approximately Americans would have a IQ score more than higher than 140
Step-by-step explanation:
1) Previous concepts
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".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
2) Solution to the problem
Let X the random variable that represent the IQ scores of the population of interest, and for this case we know the distribution for X is given by:
[tex]X \sim N(100,19)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=19[/tex]
We are interested on this probability
[tex]P(X>140)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula and the complement rule to our probability we got this:
[tex]P(X>140)=1-P(X\leq 140)=1-P(\frac{X-\mu}{\sigma}<\frac{140-\mu}{\sigma})=1-P(Z<\frac{140-100}{19})=1-P(Z<2.105)=0.0176[/tex]
And we can find this probability with the following excel code:
"=1-NORM.DIST(2.105,0,1,TRUE)"
This number 0.0176 represent the proportion of Americans that present a score higher than 140.
And now since we ar einterested on the approximate number of people in the United States (assuming a total population of 280,000,000) with an IQ higher than 140, we just need to do this:
Number =0.0176*280000000=4928000 approximately Americans would have a IQ score more than higher than 140
Suppose that nine bats was used. For each trail, the zoo keeper pointed to one of two "feeders" Suppose that the bats went to the correct feeder (the one that the zoo keeper pointed at) 7 times. Find the 95% confidence interval for the population proportion of times that the bats would follow the point.
A. (0.59, 1.05)
B. (0.44,0.94)C. (0.51, 1.0)
Answer: B. (0.44,0.94)
Step-by-step explanation:
Given : Number of observations : n = 9
Number of successes : x = 7
Let p be the population proportion of times that the bats would follow the point.
Since the sample size is small , so we use plus four confidence interval for p.
Plus four estimate of p=[tex]\hat{p}=\dfrac{\text{No. of successes}+2}{\text{No. of observations}+4}[/tex]
[tex]=\dfrac{7+2}{9+4}\approx0.69[/tex]
By z-table , the critical value for 95% confidence level : z* = 1.96
Then, the 95% confidence interval for the population proportion of times that the bats would follow the point. will be :
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}}[/tex] , where N= 13
[tex]0.69\pm (1.96)\sqrt{\dfrac{0.69(1-0.69)}{13}}[/tex]
[tex]0.69\pm (1.96)\sqrt{0.0163862084615}[/tex]
[tex]0.69\pm (1.96)(0.128008626512)[/tex]
[tex]\approx0.69\pm 0.25=(0.69-0.25,\ 0.69+0.25)[/tex]
[tex](0.44,\ 0.94)[/tex]
Hence, the 95% confidence interval for the population proportion of times that the bats would follow the point = [tex](0.44,\ 0.94)[/tex]
Thus the correct answer is B. (0.44,0.94)
To find the 95% confidence interval for the population proportion, use the formula CI = p ± z * √((p(1-p))/n), where p is the sample proportion, z is the z-score, and n is the sample size. Substituting values, the 95% confidence interval is approximately (0.685, 0.869).
Explanation:To find the 95% confidence interval for the population proportion, we can use the formula:
CI = p ± z * √((p(1-p))/n)
where p is the sample proportion, z is the z-score for the desired confidence level, and n is the sample size.
In this case, the sample proportion is 7/9 and n is 9. Since we want a 95% confidence interval, the z-score is approximately 1.96.
Substituting these values into the formula:
CI = (7/9) ± 1.96 * √(((7/9)(2/9))/9)
CI = 0.777 ± 1.96 * √(0.123/9)
CI ≈ 0.777 ± 1.96 * 0.047
CI ≈ (0.777 - 0.092, 0.777 + 0.092)
CI ≈ (0.685, 0.869)
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What is a "confound" in an experimental study? Group of answer choices
a. an extra variable that could explain result differences between groups
b. the tendency of a control group to mimic the experimental group
c. a control group that receives special attention
d. an experimental group that fails to respond to treatment
Answer:
The correct option is option a. an extra variable that could explain result differences between groups.
Step-by-step explanation:
Confusion variables, also called third variables, are variables that the investigator did not control or did not eliminate and that damage the internal validity of an experiment, that is, the degree to which the results are valid.
The dependent and independent variables are the two main variables of any experiment or investigation. The independent is one that changes or is controlled to study its effects on the dependent variable. The dependent is that variable that is investigated and measured.
Confusion variables may cause the investigator to analyze the results incorrectly. The results may show a false correlation between dependent and independent variables.
So the correct option is option a. an extra variable that could explain result differences between groups.
This is because this additional variable that was not controlled or taken into account during the experimental study modifies the result. And the other groups that conducted the experiment and took into account the variable will have different results because they take into account other independent variables.
Given that events A and B are independent and that P(A) = 0.8 and P(B|A) = 0.4, then P(A and B) = 0.32.
a. True
b. False
Answer: Option 'a' is correct.
Step-by-step explanation:
Since we have given that
P(A) = 0.8
P(B|A) = 0.4
Since A and B are independent events.
Since P(B|A) is given, we will use the formula for "conditional probability":
So, [tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\0.4=\dfrac{P(A\cap B)}{0.8}\\\\0.4\times 0.8=P(A\cap B)}\\\\0.32=P(A\cap B)[/tex]
Hence, it is true.
Therefore, Option 'a' is correct.
Find at least ten solutions to the linear equation 1/2x + y = 5, and plot the points on a coordinate plane. What shape is the graph of the linear equation taking?
The linear equation 1/2x + y = 5, we can choose values for x and solve for y. When we plot the points on a coordinate plane, we get a line graph.
Explanation:To find solutions to the linear equation 1/2x + y = 5, we can arbitrarily choose values for x and solve for y.
Let's choose x = 0:
1/2(0) + y = 5
y = 5
So one solution is (0, 5).
Now let's choose x = 2:
1/2(2) + y = 5
1 + y = 5
y = 4
Another solution is (2, 4).
We can continue this process and find more solutions:
x = 4, y = 3
x = 6, y = 2
x = 8, y = 1
x = 10, y = 0
x = -2, y = 6
x = -4, y = 7
x = -6, y = 8
x = -8, y = 9
x = -10, y = 10
These are ten solutions to the equation.
If we plot these points on a coordinate plane, we will see that they all lie on a straight line.
Therefore the shape of the graph is a line. The equation represents a linear relationship between x and y.
The scale drawing has a scale of 1/2 in: 8 mi. Find the length on the drawing for 2 in Please answer asap
Answer:
32 mi
Step-by-step explanation:
Solve using proportions.
[tex]\frac{\frac{1}{2}in}{8 mi} =\frac{2in}{y}[/tex]
Find the scale factor (how to get from left to right)
To get from left numerator to right numerator, multiply by 4.
(1/2) X 4 = 2
The scale factor is 4.
Multiply the left denominator by the scale factor to get "y".
8 mi X 4 = 32 mi
Therefore 2 inches represent 32 miles.
Provide in simplest form
12% of 40
The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of a bridge is 1270 ft long and 157 ft high. The parabola y=0.00039x^2 gives a good fit to the shape of the cables, where IxI less than of equal to 635, and x and y are measured in feet. Approximate the length of the cables that stretch between the tops of the two towers.
Answer:
Step-by-step explanation:
Given
span of bridge [tex]L=1270\ ft[/tex]
height of span [tex]h=157\ ft[/tex]
Equation of Parabola
[tex]y=0.00039x^2[/tex]
[tex]|x|<635[/tex] i.e.
[tex]-635<x<635[/tex]
[tex]\frac{dy}{dx}=2\times 0.00039[/tex]
length of Arc[tex]=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}[/tex]
[tex]=\int_{-635}^{635}\sqrt{1+(\frac{dy}{dx})^2}[/tex]
[tex]=\int_{-635}^{635}\sqrt{1+(0.00078x)^2}[/tex]
[tex]=2\times \int_{0}^{635}\sqrt{1+(0.00078x)^2}[/tex]
[tex]=2\times (660.08)[/tex]
[tex]=1320.16\ ft[/tex]
The approximate length of the cables is approximately 4534.24 feet.
To approximate the length of the cables that stretch between the tops of the two towers of the suspension bridge, we can use the integral calculus to find the length of the curve defined by the equation [tex]\(y = 0.00039x^2\).[/tex]
The formula for finding the length of a curve between two points [tex]\([a, b]\)[/tex] is given by the integral:
[tex]\[L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\][/tex]
Where:
- L is the length of the curve.
- a and b are the x-coordinates of the two points between which we want to find the length.
- f(x) is the function representing the curve.
- f'(x) is the derivative of the function.
In this case, we want to find the length of the cables between the two towers, which corresponds to the x-values from -635 to 635 since the width of the bridge is 1270 feet. The curve is defined by [tex]\(y = 0.00039x^2\)[/tex], so:
- a = -635
- b = 635
- [tex]\(f(x) = 0.00039x^2\)[/tex]
- [tex]\(f'(x) = 2 \cdot 0.00039x\)[/tex]
Now, let's calculate the length:
[tex]\[L = \int_{-635}^{635} \sqrt{1 + (2 \cdot 0.00039x)^2} \, dx\][/tex]
[tex]\[L = \int_{-635}^{635} \sqrt{1 + 0.0001536x^2} \, dx\][/tex]
Now, we can evaluate this integral:
[tex]\[L = \int_{-635}^{635} \sqrt{1 + 0.0001536x^2} \, dx \approx 4534.24\][/tex]
So, the approximate length of the cables that stretch between the tops of the two towers of the suspension bridge is approximately 4534.24 feet.
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At the local racetrack, the favorite in a race has odds 3:2 of losing. What is the probability that the favorite wins the race?
a. 0.2
b. 0.67
c. 0.6
d. 0.4
Answer:
0.40
Step-by-step explanation:
Given that at the local racetrack, the favorite in a race has odds 3:2 of losing
Here instead of probability odds are given.
Odds of losing = 3/2
Hence Probability of losing = [tex]\frac{3}{3+2} \\=\frac{3}{5} \\=0.6[/tex]
Probability that the favourite wins the race will be the probability for the event which is complement of losing the game.
Hence
Probability that the favourite wins the race will be the probability
= 1- 0.6
=0.40
Probability that favorite wins the race is 0.6
Given that;Favorite in a race has odds = 3:2
Find:Probability that favorite wins the race
Computation:Probability that favorite wins the race = 3 / [3 + 2]
Probability that favorite wins the race = 3 / 5
Probability that favorite wins the race = 0.6
Option "C" is the correct answer to the following question.
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Assume that it takes a college student an average of 5 minutes to find a parking spot in the main parking lot. Assume also that this time is normally distributed with a standard deviation of 2 minutes. Find the probability that a randomly selected college student will take between 2 and 6 minutes to find a parking spot in the main parking lot.
Answer:
0.624 is the probability that a randomly selected college student will take between 2 and 6 minutes to find a parking spot in the main parking lot.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 5 minutes
Standard Deviation, σ = 2 minutes
We are given that the distribution of time is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(student will take between 2 and 6 minutes )
[tex]P(2 \leq x \leq 6) = P(\displaystyle\frac{2 - 5}{2} \leq z \leq \displaystyle\frac{6-5}{2}) = P(-1.5 \leq z \leq 0.5)\\\\= P(z \leq 0.5) - P(z < -1.5)\\= 0.691 - 0.067 = 0.624 = 62.4\%[/tex]
[tex]P(2 \leq x \leq 6) = 62.4\%[/tex]
0.624 is the probability that a randomly selected college student will take between 2 and 6 minutes to find a parking spot in the main parking lot.
The PCB concentration of a fish caught in Lake Michigan was measured by a technique that is known to result in an error of measurement that is normally distributed with a standard deviation of .08 ppm (parts per million). Suppose the results of 10 independent measurements of this fish are 11.2, 12.4, 10.8, 11.6, 12.5, 10.1, 11.0, 12.2, 12.4, 10.6(a) Give a 95 percent confidence interval for the PCB level of this fish.(b) Give a 95 percent lower confidence interval.(c) Give a 95 percent upper confidence interval.
Final answer:
The 95% confidence interval for the PCB level in the fish is (11.4228, 11.5372) ppm. The 95% lower confidence interval is 11.4228 ppm and the 95% upper confidence interval is 11.5372 ppm, based on both the t-distribution and the provided sample data and standard deviation.
Explanation:
To find the 95% confidence interval, lower confidence interval, and upper confidence interval of the PCB level in a fish from Lake Michigan, based on 10 measurements and a standard deviation of 0.08 ppm, we first need to calculate the sample mean and then apply the appropriate formulas.
To calculate the mean PCB concentration (μ), we sum all the values and divide by the number of measurements (n=10):
μ = (11.2 + 12.4 + 10.8 + 11.6 + 12.5 + 10.1 + 11.0 + 12.2 + 12.4 + 10.6) / 10 = 114.8 / 10 = 11.48 ppm.
For calculating the confidence intervals, we use the t-distribution since the sample size is small. We need the t-value for 9 degrees of freedom (n-1) at the 95% confidence level which, assuming it is approximately 2.262 (values differ slightly depending on the t-distribution table used).
The standard error (SE) is calculated using the sample standard deviation (s) and the square root of the number of measurements: SE = s/sqrt(n) = 0.08/sqrt(10) = 0.0253 ppm.
The 95% confidence interval is given by:
CI = μ ± (t-value * SE)
CI = 11.48 ± (2.262 * 0.0253)
CI = 11.48 ± 0.0572
CI = (11.4228, 11.5372) ppm
The 95% lower confidence interval is the mean minus the product of the t-value and SE:
LCI = μ - (t-value * SE)
LCI = 11.48 - (2.262 * 0.0253)
LCI = 11.4228 ppm
The 95% upper confidence interval is the mean plus the product of the t-value and SE:
UCI = μ + (t-value * SE)
UCI = 11.48 + (2.262 * 0.0253)
UCI = 11.5372 ppm