Ask question

Ask question

All categories

3 years ago

15

HURRY 40 POINTS !!!!!!!!! A person runs 1/4 of a mile in 1/32 of an hour. Compute the unit rate by setting up a complex fraction

and simplifying.
A) 4 miles per hour
B) 6 miles per hour
C) 8 miles per hour
D) 12 miles per hour

2 answers:

hram777 [196]3 years ago

5 0

I am not sure, but the best answer should be 8 miles per hour

son4ous [18]3 years ago

5 0

c or 8 miles per hour and it's correct becuase I did with calculations

You might be interested in

What is the slope of the line that passes through (-5,-2) and ( 7, -2)

denis23 [38]

｡☆✼★ ━━━━━━━━━━━━━━ ☾

slope = difference in y / difference in x

Sub the values in:

slope = (-2- -2) / (7 - -5)

slope = 0 / 12

The slope is 0.

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

｡☆✼★ ━━━━━━━━━━━━━━ ☾

6 0

3 years ago

Read 2 more answers

Please help me solve this equation

elena-s [515]

Answer:

m<U = 38degrees

Step-by-step explanation:

From the given diagram, <B = <U since both triangles are similar, hence;

Hence;

2y+2 = 3y-16

2y - 3y = -16 - 2

-y = -18

y = 18

Get m<U

m<U = 3y-16

m<U = 3(18) -16

m<U = 54 - 16

m<U = 38degrees

3 0

3 years ago

Andy is alone in the wilderness and eager to get back to camp. Looking at his map (shown below), Andy sees that he might be able

BabaBlast [244]

Answer:

9.5 km

Step-by-step explanation:

8 0

3 years ago

A U.S. Travel Data Center survey of 500 adults found that 195 said they would take more vacations this year than last year. Cons

Lelechka [254]

Using the z-distribution, it is found that the 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

195 out of 500 people said they would take more vacations this year than last year, hence:

$n = 500, \pi = \frac{195}{500} = 0.39$

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.39 - 1.96\sqrt{\frac{0.39(0.61)}{500}} = 0.347$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.39 + 1.96\sqrt{\frac{0.39(0.61)}{500}} = 0.433$

The 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).

A similar problem is given at brainly.com/question/15850972

5 0

3 years ago

60 boys at a summer camp were asked to complete a survey about their favorite outdoor activity.

ikadub [295]

Answer:

48.3%.

Step-by-step explanation:

To determine the percentage of children in the summer camp who chose fishing as their favorite activity, knowing that there were 29 of 60 children in total who chose that option, the following calculation must be performed:

60 = 100

29 = X

(29 x 100) / 60 = X

2,900 / 60 = X

48.3 = X

Therefore, 48.3% of the children chose fishing as their favorite activity in summer camp.

8 0

3 years ago

Read 2 more answers