Determine the unit rate of a marathon runner who travels 5/2 miles in 1/4 hour

I really need help. Please help me!

Mumz [18]

Answer:

Simple the answer is 0

Step-by-step explanation:

The initial value is the y intercept or where the line passes though on the y axis

5 0

4 years ago

Read 2 more answers

The accompanying data represent the daily (for example, Monday to Tuesday) movement of Johnson & Johnson (JNJ) stock for

egoroff_w [7]

Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

The lower bound is of 0.484.
The upper bound is of 0.7292.

The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

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 $\frac{1+\alpha}{2}$ .

Supposing that it increases on 37 out of 61 days:

$n = 61, \pi = \frac{37}{61} = 0.6066$

95% confidence level

So $\alpha = 0.95$ , 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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 - 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.484$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 + 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.7292$

The 95% confidence interval for the proportion of days JMJ stock increases is (0.484, 0.7292), in which 0.484 is the lower bound and 0.7292 is the upper bound.

The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

A similar problem is given at brainly.com/question/16807970

4 0

3 years ago

A square garden plot measures 180 square feet. A second square garden plot measures 320 square feet. How many more feet of fence

REY [17]

The second garden plot will require 8√5 feet more fence than the first garden plot.

Further explanation:

In order to find the fence, we have to find the perimeter of both squares

So,

Area of Square 1: A1=180 square feet

Area of Square 2: A2=320 Square feet

Let x be the side of square 1:

Then,

$A_1=x^2\\180=x^2\\Taking\ square\ root\ on\ both\ sides\\\sqrt{180}=\sqrt{x^2}\\x=\sqrt{2*2*3*3*5}\\x=\sqrt{2^2*3^2*5}\\x=2*3\sqrt{5}\\x=6\sqrt{5}$

For second square:

Let y be the side of second square

$A_2=y^2\\320=y^2\\Taking\ square\ root\ on\ both\ sides\\\sqrt{320}=\sqrt{y^2}\\y=\sqrt{2*2*2*2*2*2*5}\\y=\sqrt{2^2*2^2*2^2*5}\\y=2*2*2\sqrt{5}\\y=8\sqrt{5}$