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dimulka [17.4K]

3 years ago

9

Jason plans to join a health club. The club charges $6.50 per week plus a $20 enrollment fee. If Jason has budgeted to spend no

more than $250 this year on health club expenses, how many weeks will he be able to workout at the club?
A. 20w + 6.5 ≥ 250

B. 20 + 6.5w ≥ 250

C. 20w + 6.5 ≤ 250

D. 20 + 6.5w ≤ 250

2 answers:

Ray Of Light [21]3 years ago

7 0

Take 250$ - 20$ = 230$
then 230$ divide by 6.50$ per week
you get 35,38461538461538 weeks
or about 35 weeks

Leona [35]3 years ago

6 0

The answer is B 20 +6.5w>250

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Step-by-step explanation:

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A national survey conducted in 2011 among a simple random sample of 1,507 adults shows that 56% of Americans think the Civil War

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Answer:

a) $z=4.658$

$p_v =P(z>4.65)=1-P(z$

b) Using the significance level assumed $\alpha=0.05$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is higher than 50%.

c) The 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

Step-by-step explanation:

I )Part a

1) Data given and notation

n=1507 represent the random sample taken

X represent the Americans who thinks that the Civil War is still relevant to American politics and political life

$\hat p$ estimated proportion of Americans who thinks that the Civil War is still relevant to American politics and political life in the sample

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha=0.05$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of Americans who thinks that the Civil War is still relevant to American politics and political life

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =1507*0.5=753.5>10$

$n(1-p_o)=1507*(1-0.5)=753.5>10$

3) Calculate the statistic

The statistic is calculated with the following formula:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

On this case the value of $p_o=0.5$ is the value that we are testing and n = 1507.

$z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1507}}}=4.658$

The p value for the test would be:

$p_v =P(z>4.65)=1-P(z$

II) Part b

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

Based on the alternative hypothesis the p value would be given by:

$p_v =P(z>4.65)=1-P(z$

Using the significance level assumed $\alpha=0.05$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is higher than 50%.

III) Part c

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.56 - 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.527$

$0.56 + 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.593$

And the 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

And this result agrees with the result of part b, since the interval not contains the value of 0.5 we can conclude that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life it's higher than 0.5 at 90% of confidence.

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Answer:

a) In the explanation.

b) P=0.09714

c) P=0.0866

d) p=0.208

Step-by-step explanation:

a) The sampling distribution will have this properties:

1 - The sampling distribution of a proportion is equal to the binomial distribution.

2 - The mean of the sampling distribution will be equal to the proportion of the population (0.328).

$\mu=\pi=0.328$

3 - The standard deviation will be equal to the square root of the product of the proportion and its complementary, divided by the sample size.

$\sigma_s=\sqrt{\pi(1-\pi)/n}=\sqrt{0.328*(1-0.328)/25}=0.094$

B) The fraction with sample proportion of 0.45 or higher is

$z=(p-\pi)/\sigma=(0.45-0.328)/0.094=1.298\\\\P(p\geq0.45)=P(z\geq 1.298)=0.09714$

C) The fraction with sample proportion of 0.20 or lower is

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AVprozaik [17]

Answer: (x - 1)(x + 1)^2(x - 4)^3

Step-by-step explanation:

First, remember that:

in expressions like: (x - b)^n

b is the value of x where the graph intersects the x-axis.

n can represent:

n = 1, the line just goes through the x-axis

n = 2, the line may change the direction (an inflection point), and touch the x-axis in one point.

n = 3, the line may have two inflection points when it intersects the x-axis.

Then we have the expression:

(x-b)(x-c)^2(x-d)^3

b is in the linear part, the graph crosses the x-axis linearly in x = 1.

c is in the quadratic part, the graph crosses the x-axis with one point of inflection at x = -1.

d is in the cubic part, the graph crosses the x-axis with two inflections in x = 4.

Then we can writhe the polynomial as:

f(x) = (x-1)(x-(-1))^2(x-4)^3 = (x - 1)(x + 1)^2(x - 4)^3

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