Help!!!!! Solve the distance??

Whats the area of a rectangle 17ft 4ft

Katena32 [7]

Answer:

68

Step-by-step explanation:

Multiply the numbers to get the area

7 0

2 years ago

Read 2 more answers

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

A 5-pound box of raspberries costs $27.20. What is the price per ounce?<br> $

ASHA 777 [7]

Answer:

$0.34 per ounce

Step-by-step explanation:

A 5 pound box can be changed to ounces. There are 16 ounces in a pound. So we have 5 groups of 16.

5 × 16 is 80 ounces.

There are 80 ounces in 5lbs.

To get a price per ounces, divide.

dollars/ounces

= 27.20/80

= 0.34 dollars/oz.

This is 34cents per ounce.

6 0

1 year ago

A student has $1 bills and $5 bills in his wallet. He had a total of 15 bills that are worth $47. How many of each type does he

Mrrafil [7]

Total number of $1 bills=7

Total number of $5 bills=8

Further explanation:

Given

Let x be the number of $1 bills

and

y be the number of $5 bills

Then according to the given statements

x+y=15

x+5y=47

From first equation:

$x+y=15\\x=15-y$

Putting x=15-y in second equation

$x+5y=47\\15-y+5y=47\\4y=47-15\\4y=32\\\frac{4y}{4}=\frac{32}{4}\\y=8$

Putting y=8 in first equation

$x+8=15\\x=15-8\\x=7$

Hence,

Total number of $1 bills=7

Total number of $5 bills=8

Keywords: Linear equations, Substitution method

Learn more about linear equations at:

brainly.com/question/2367554

brainly.com/question/2977815

#LearnwithBrainly

4 0

3 years ago

A store buys a jacket for $20.00 and sells it to customers for 80% more than that. What is the selling price of the jacket

Tamiku [17]

38 because 80% of 20 is 18 and then add the 2

5 0

2 years ago