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The length of a rectangle is four times its width. If the perimeter of the rectangle is 50 cm, find its area.

Answers

Answer 1

Answer:

width=25/3

Length=75/3

Step-by-step explanation:

Given

Perimeter=50

Let width=x

Length=4x

So perimeter =2(length+width)

2(length+width)=50

2(X+2x)=50

3x=25

X=25/3

So width=25/3

Length=75/3

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Simplify the polynomial

Answers

The polynomial after simplification is obtained as $5 x^(2)$ + 9x + 3.

What issimplification?

In mathematics, reducing an expression, fraction, or problem to a simpler form is referred to as simplification. With calculations and solution, the problem is made simple. Make something simpler by simplifying it.

We are given a polynomial as

5x - 5 + $11 x^(2)$ + 4x - $6 x^(2)$ + 8

Now, in order to simplify the given polynomial, we will combine the like terms of the polynomial.

On combining the like terms, we get

$5 x^(2)$ + 9x + 3

Hence, the polynomial after simplification is obtained as $5 x^(2)$ + 9x + 3.

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You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you survey? Assume that you want to be 99% confident that the sample percentage is within 5.5 percentage points of the true population percentage. Complete parts (a) and (b) below. a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. nequals 549 (Round up to the nearest integer.) b. Assume that a prior survey suggests that about 34% of air passengers prefer an aisle seat. nequals nothing (Round up to the nearest integer.)

Answers

Answer:

a) $n=(0.5(1-0.5))/(((0.055)/(2.58))^2)=550.116$

And rounded up we have that n=551

b) $n=(0.34(1-0.34))/(((0.055)/(2.58))^2)=493.78$

And rounded up we have that n=494

Step-by-step explanation:

Previous concept

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

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_(\alpha/2)=-2.58, t_(1-\alpha/2)=2.58$

Part a

The margin of error for the proportion interval is given by this formula:

$ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\hat p (1-\hat p))/(((ME)/(z))^2)$ (b)

We can assume that $\hat p =0.5$ since we don't know prior info. And replacing into equation (b) the values from part a we got:

$n=(0.5(1-0.5))/(((0.055)/(2.58))^2)=550.116$

And rounded up we have that n=551

Part b

$n=(0.34(1-0.34))/(((0.055)/(2.58))^2)=493.78$

And rounded up we have that n=494

Final answer:

To determine the required sample size for the survey, we can use a sample size formula based on the desired confidence level and margin of error. If nothing is known about the passenger preferences, a sample size of 549 would be needed. If a prior survey suggests a certain proportion, the sample size can be calculated using the known proportion.

Explanation:

In order to determine the number of randomly selected air passengers that must be surveyed, we need to calculate the required sample size for a desired confidence level and margin of error.

a. If nothing is known about the percentage of passengers who prefer aisle seats, we can use a sample size formula given by n = (Z^2 * p * q) / E^2, where Z is the z-score corresponding to the desired confidence level, p and q are the estimated proportions for aisle seat preference and non-aisle seat preference respectively, and E is the desired margin of error. Since a confidence level of 99% and a margin of error of 5.5% are specified, we can round up the sample size to 549.

b. If a prior survey suggests that about 34% of air passengers prefer an aisle seat, we can use the same sample size formula but with the known proportion p = 0.34. We do not have information about the non-aisle seat preference, so we cannot determine the required sample size.

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PLSSS CAN SOMEONE HELP MEE, my friend can't figure this out , and the assignment is already overdue , pls someone help :( this is the first part of it , it also has a B part

Answers

Answer:

um 1630 divided by 96 and divided by 3equals the answer ? yes yes i think

Step-by-step explanation:

Quan likes to exercise by running every day. On average, he runs 5 miles each day. This varies by about 0.3 mile. What are the maximum and minimum numbers of miles Quan is expected to run each day?

Answers

Answer:

Quan runs a maximum of 5.3 miles and a minimum of 4.7 miles every day.

Step-by-step explanation:

Let x= the actual measurement.

|actual − ideal|≤ tolerance

Use an absolute value inequality to express this situation.

|x−5|≤0.3

Rewrite as a compound inequality.

−0.3≤x−5≤0.3

Solve the inequality.

4.7≤x≤5.3

Answer the question. Quan runs a maximum of 5.3 miles and a minimum of 4.7 miles every day.

Use the table to find the ratio. Enter the ratio as a fraction in simplest form. 9 white socks. 6 black socks. 7 blue socks. 2 brown socks. What is the ratio of white socks to brown socks?

Answers

Answer:

Step-by-step explanation:

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from the table, the ratio of white socks to brown socks is 9.4 select two ratios equilavent to the ratio of white socks to brown socks

The following table shows the number of snow days each school district in Mill County had last winter. School District District 200200200 District 211211211 District 221221221 District 231231231 District 241241241 Number of snow days 666 888 333 222 666 Find the mean absolute deviation (MAD) of the data set. snow days

Answers

The number of snowdays of District 241 are 4.

What is Statistics?

Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data.

We have to find the number of snowdays of District 241

School District District201 District211 District221 District231 District 241

Number of 4 8 3 6 ?

snow days

Mean of snow days is 5.

Mean =Sum of observations/Number of observations

5=4+8+3+6+x/5

25=21+x

Subtract 21 from both sides

x=4

Hence, the number of snow days of District 241 are 4.

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Answer:

DIstrict 241 had 4 snow days.

Step-by-step explanation:

5 * 5 = 25

Add the ones you know

4 + 8 + 3 + 6 = 21

Then

25 - 21 = 4

So District 241 had 4 snow days.

I know this answer is 100% correct. I answered it correctly. This problem wasn't that hard. Let me know if you need help with anything else.

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