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snow_tiger [21]

3 years ago

13

The perimeter of a rectangle is 80 feet. Find the dimensions if the length 5 feet longer than four times the width. Then find th

e area of the rectangle.

1 answer:

nydimaria [60]3 years ago

3 0

Answer:

l=33 feet

w=7 feet

Area= 231 feet

Step-by-step explanation:

Solution in Picture is attached

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1A technical magazine stated that the mean retail cost of all models of a particular laptop was $600. A random sample of 10 stor

My name is Ann [436]

Answer:

Yes

Step-by-step explanation:

Given,

Population mean Z = 600

Sample mean, µ = 586.50

Size of the sample, n = 10

Standard deviation, σ = 26.77

The testing hypothesis is,

Null hypothesis = µ<600

Alternative hypothesis = µ≥600

The test statistics is the calculated value of Z,

(Z- µ)/(σ/√n)

= (600-586.50)/(26.77/√10)

= (13.5*3.1623)/26.77

= 42.691/26.77

= 1.5947

And at 5% level of significance, the table value of Z = 1.644

Here, calculated Z<Table value of Z, so we accept the null hypothesis.

So, the null hypothesis is accepted. The mean value of the laptop is less than 600.

3 0

3 years ago

Given that ∠2 ≅ ∠6.

insens350 [35]

Answer:

Converse of the alternate exterior angle theorem

Step-by-step explanation:

just took the test ;)

3 0

3 years ago

Read 2 more answers

Phillip wanted to leave a 15% tip. Phillip's bill was $36.00. How much will he leave for a tip?

Anastaziya [24]

Your answer would be letter B. $5.40

4 0

2 years ago

Read 2 more answers

Question 3 (1 point)

den301095 [7]

Answer: Pay 250 per month until it’s paid off

4 0

2 years ago

A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A ran

mart [117]

Answer:

a) The 95% confidence interval would be given (0.070;0.121).

b) "increase the sample size n"

"decrease za/2 by decreasing the confidence"

Step-by-step explanation:

Notation and definitions

$X=48$ number of vehicles classified as sports utility.

$n=500$ random sample taken

$\hat p=\frac{48}{500}=0.096$ estimated proportion of vehicles classified as sports utility vehicles.

$p$ true population proportion of vehicles classified as sports utility vehicles.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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 population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

(a) Use a 95% confidence interval to estimate the proportion of sports utility vehicles in California. (Round your answers to three decimal places.)

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 95% confidence interval the value of $\alpha=1-0.5=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.096 - 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.070$

$0.096 + 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.121$

And the 95% confidence interval would be given (0.070;0.121).

We are confident (95%) that about 7.0% to 12.1% of vehicles in California are classified as sports utility .

(b) How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers. Select all that apply.)

For this case we just have two ways to increase the accuracy one is "increase the sample size n" since if we have a larger sample size the estimation would be more accurate. And the other possibility is "decrease za/2 by decreasing the confidence" because if we decrease the confidence level the interval would be narrower and accurate

7 0

2 years ago