Which expression is an equivalent expression of 12x + 10 + 4y?

To plan the budget for next year a college must update its estimate of the proportion of next year's freshmen class that will ne

Naily [24]

Answer:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

$\hat p=\frac{67}{150}=0.447$ estimated proportion of applicants who request financial aid

$p_o=0.35$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 0.35.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

Statistical decision

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.

The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

5 0

1 year ago

Mrs. Start's class paid $5.60 for 4 5 yd of lace for an art project. Later, the class decided to buy another 7 3 4 yd of the sam

Damm [24]

The answer to your question, is 54.25.

3 0

1 year ago

Pavi has a credit line of $8,000 on his credit card. Review the summary of his credit card statement. How much credit does Pavi

jekas [21]

Answer:

A. $3992.50

Step-by-step explanation:

Credit line - $8,000.00

Summary

Previous Balance Payments - $2,500.00
Credits - $1,500.00
New Purchases - $3,000.00
Finance Charge - $7.50
New Balance - $4,007.50

Available credit= credit line - new balance = $8000 - $4007.50= $3992.50

8 0

1 year ago

According to a survey, 10% of Americans are afraid to fly. Suppose 1100 Americans are sampled. a) What is the probability that 1

JulijaS [17]

Answer:

a) 14.46%

b) 0.00%

c) 1.54%

Step-by-step explanation:

According to the survey, 10% of Americans are afraid to fly.

This means $p=0.10$ and $q=1-0.10=0.90$ .

If 1100 Americans are sampled, then the sample size is $n=1100$ .

The mean of the distribution is $\mu=np$ .

This means $\mu=1100\times 0.10=110$

The standard deviation is $\sigma=\sqrt{npq}$

We substitute the values to get:

$\sigma=\sqrt{1100\times 0.1\times 0.9}=9.95$

a) We want to find the probability that 121 or more Americans in the survey are afraid to fly.

We first apply the continuity correction factor to get: $P(X\ge121)=P(X\:>\:121-0.5)\\P(X\ge121)=P(X\:>\:120.5)$

We now convert to Z-scores to get:

$P(X\:>\:120.5)=P(z\:>\:\frac{120.5-110}{9.95})=1.06\\$

From the standard normal distribution table P(z>1.06)=0.1446

As a percentage, the probability is 14.46%

b) We want to find the probability that 165 or more Americans in the survey are afraid to fly.

We apply the CCF to get:

$P(X\ge165)=P(X\:>\:165-0.5)\\P(X\ge165)=P(X\:>\:164.5)$

We convert to z-scores:

$P(X\:>\:164.5)=P(z\:>\:\frac{164.5-110}{9.95})=5.48$

From the normal distribution, P(z>164.5)=0

c) First, 8% of 1100 is 88.

We want to find the probability that 88 or less Americans in the survey are afraid to fly.

We apply the CCF to get:

$P(X\le88)=P(X\:$

We convert to z-scores:

$P(X\:$

From the normal distribution, P(z>\:-2.16)=0.0154

As a percentage, we get 1.54%

3 0

2 years ago

The probability of flowering plants bearing flowers in January is given the table. If a plant has flowered in January, what is t

ololo11 [35]

Answer:

Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

Step-by-step explanation:

We know that:

Probability =Total favourable outcomes/ Total possible outcomes

Probability of flowering plants bearing flowers in January is given in the table. It means that out of 100 peony 70 peony flowers in January.

There are five kinds of flowers.

We have to select peony and find it's Probability.

As all flowers are independent, their flowering phenomenon does not depend on each other.

So,Probability of a flowering plant bearing flowers in January if it is a peony= 70 %

7 0

1 year ago

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