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3 years ago

A student spent less than $26 on a binder and 5

Mathematics

2 answers:

Cerrena [4.2K]3 years ago

6 0

Answer:

$4.20.

Step-by-step explanation:

My equation is:

5c+$4=26

To solve I do:

$26-$4=$22

$22 is the total amount of the composition books.

$22 ÷ 5 = $4.20

The maximum cost of each composition book is $4.20.

Ray Of Light [21]3 years ago

5 0

22/5=4.4
26>c is less or equal to 4.39

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Juana wants to use the numbers 8, 6, 3,

andrezito [222]

She can create 16 because there’s 4 numbers times the 4 digit code, she have 16 different possibilities of #s

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3 years ago

Is 8 a solution to the equation below? Yes or No?<br><br><br> 3(x - 2) + 4 = 22

Simora [160]

To see if 8 is the solution to the equation, 3 (x - 2) + 4 = 22, you must replace x with 8 and solve using the rules of PEMDAS (Parentheses, Exponent, Multiplication, Division, Addition, Subtraction) REMEMBER: IF NOT APPLICABLE TO THE EQUATION YOU MAY SKIP THAT STEP IN PEMDAS . If the sides are equal to each other then 8 IS a solution.

3(8 - 2) + 4 = 22

Parentheses...

3(8 - 2) + 4 = 22

8 - 2 <<<In the parentheses so you must solve first

so...

3(6) + 4 = 22

Multiplication...

3(6) + 4 = 22

3 * 6

so..

18 + 4 = 22

Addition...

18 + 4 = 22

18 + 4

so...

22 = 22

Are they equal to each other? YES!!!

8 is a solution to the equation 3 (x - 2) + 4 = 22

Hope this helped!

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3 years ago

he American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 45 w

Sonja [21]

Answer:

The 85% confidence interval is (0.2187, 0.2613).

Step-by-step explanation:

We have a large sample of size n = 830 Americans over 45. We have observed that 631 people don't smoke. Let p be the population proportion of Americans over 45 who smoke. An point estimate of p is $\hat{p} = 199/830 = 0.24$ and an estimate of the standard deviation of $\hat{p}$ is $\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{(0.24)(0.76)/830}$ = 0.0148. Therefore, because we have a large sample, the 85% confidence interval for the population proportion is given by $\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n} = 0.24\pm z_{0.15/2}0.0148 = 0.24\pm z_{0.075}0.0148$ where $z_{0.075}$ is the 7.5th quantile of the standard normal distribution, i.e., -1.4395. Therefore the 85% confidence interval is $0.24\pm (1.4395)(0.0148)$ or equivalently (0.2187, 0.2613).