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2 years ago

HELP PLEASE!!!! I REALLY NEED HELP

Mathematics

1 answer:

Free_Kalibri [48]2 years ago

3 0

Answer:

CDW stocks a broad selection of office furniture, supplies, and accessories from computer stands to cleaning wipes, paper shredders and monitor stands.

Step-by-step explanation:

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draw three rectangles.Then draw to show halves,thirds,and fourths. Write about each whole that you have drawn

Yuri [45]

Draw a rectangle, draw a line in middle. This is a halve.
Draw another rectangle, draw two lines. This is thirds.
Draw another rectangle, draw two lines but in a "+" type of way.. This is fourths.

6 0

3 years ago

A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measurin

Ratling [72]

Answer:

The standard deviation of weight for this species of cockroaches is 4.62.

Step-by-step explanation:

Given : A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measuring the weights of many of these cockroaches, a lab assistant reports that 14% of the cockroaches weigh more than 55 grams.

To find : What is the approximate standard deviation of weight for this species of cockroaches?

Solution :

We have given,

Mean $\mu=50$

The sample mean x=55

A lab assistant reports that 14% of the cockroaches weigh more than 55 grams.

i.e. P(X>55)=14%=0.14

The total probability needs to sum up to 1,

$P(X\leq 55)=1-P(X>55)$

$P(X\leq 55)=1-0.14$

$P(X\leq 55)=0.86$

The z-score value of 0.86 using z-score table is z=1.08.

Applying z-score formula,

$z=\frac{x-\mu}{\sigma}$

Where, $\sigma$ is standard deviation

Substitute the values,

$z=\frac{x-\mu}{\sigma}$

$1.08=\frac{55-50}{\sigma}$

$1.08=\frac{5}{\sigma}$

$\sigma=\frac{5}{1.08}$

$\sigma=4.62$

The standard deviation of weight for this species of cockroaches is 4.62.

4 0

3 years ago

I have another question I'm struggling with. How do I solve to find the missing angle?

snow_lady [41]

Answer:

$23465459544 + 44492711 = 597595855122256.56114163174112113 \leqslant \leqslant \geqslant yhy \times \frac{?}{?}kwwkjuujsjkoodji \beta \pi \beta \cos(2216 {59 \times \tim3.5es - = 6 \\ \\ 53 \times }^{2} )$

5 0

3 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

3 years ago

Prestige car rentals charge $44 per day plus 6 cents per mile to rent a mid sized vehicle. Gateway auto charges $35 per day plus

bulgar [2K]

Answer:

300 miles

Step-by-step explanation:

Let

x ----> the number of miles

y ----> the total cost for rent a car

we know that

The equation of the line in slope intercept form is equal to