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3 years ago

Please help ASAP I only got 19 mins left

Mathematics

1 answer:

Gala2k [10]3 years ago

5 0

Answer:

Step-by-step explanation:

you didnt send anything

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Type the expression that results from the following series of steps start with 5 and subtract p

azamat

Answer:

5-p

Step-by-step explanation:

8 0

3 years ago

Please help, I think I know the answer am I correct? +90 + -77 <br><br> I got= -13

alexandr1967 [171]

The correct answer would be +13 since 90 is positive and the greatest number

6 0

4 years ago

The base of a box is a rectangle. The width of the box is half its length. The height of the box is 0.5m. Find the volume of the

Ivan

Let x be the length and y be the width.
We know already
$y= \frac{x}{2}$
Area of the base $xy$
Area of the sides: $(x+y)0.25$
We get the equation $xy=0.25(x+y)-1.08$
We will solve the system:
$y= \frac{x}{2}\\xy=0.5(x+y)-1.08$
The above system has no real solution, please check your values.

3 0

3 years ago

Read 2 more answers

What is the simplest form of 5(x-30-3/9(2x+4)

Alex17521 [72]

Answer:

10x^2 -276.67x - 593.34

Step-by-step explanation:

is it like that ?

5(x-30-3/9)(2x+4)

(5x- 148.335)(2x+4)

10x^2 -296.67x +20x -593.34

10x^2 -276.67x - 593.34

3 0

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).

7 0

3 years ago