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1 year ago

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Paul joins a gym that has an initial membership fee and a monthly cost. He pays a total of $295 after three months, and $495 aft

er eight months.
a) Identify the monthly cost.
b) What is the initial membership fee?
c) Write an equation for the total cost C as a function of the number of months, m. (so C is your y and m is your x)
d) Find how much Paul will have paid after one year.

Please give me answer.

1 answer:

Harrizon [31]1 year ago

6 0

Answer:

a

Step-by-step explanation:

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Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

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1 year ago

When designing the jumps of a BMX racetrack, the jumps involve a launch ramp and a landing ramp. What are some facts about the p

zaharov [31]

The second third and last one

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Personnel tests are designed to test a job applicant's cognitive and/or physical abilities. A particular dexterity test is adm

Elza [17]

Answer:

26.11% of the test scores during the past year exceeded 83.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

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:

It is known that for all tests administered last year, the distribution of scores was approximately normal with mean 78 and standard deviation 7.8. This means that $\mu = 78, \sigma = 7.8$ .

Approximately what percentatge of the test scores during the past year exceeded 83?

This is 1 subtracted by the pvalue of Z when $X = 83$ . So:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{83 - 78}{7.8}$

$Z = 0.64$

$Z = 0.64$ has a pvalue of 0.7389.

This means that 1-0.7389 = 0.2611 = 26.11% of the test scores during the past year exceeded 83.

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1 year ago

Order 93% nine tenths 0.099 0.915% and 0.092 from least to greatest

andrew11 [14]

1. 0.915% (0.00915)
2. 0.092 (9.2%)
3. 0.099 (9.9%)
4. 9/10 (.9 or 90%)
5. 93% (.93)

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2 years ago

Mr. Li records the measures of the lengths of his students’ handprints. The lengths, in centimeters, are shown in the table.

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Answer:

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Step-by-step explanation:

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