Anybody know #4 I need it.

Find the interquartile range for the data. {50, 46, 56, 55, 54, 51, 45, 50, 47}

levacccp [35]

Answer:

51

Step-by-step explanation:

I think the answer is 51 because you have to find the number between the first three to the middle

6 0

3 years ago

Solve for X 1) 8(1 + 5x) + 5 = 13 + 5x

marusya05 [52]

Answer:

X=0

Step-by-step explanation:

$8(1 \times 5x) + 5 = 13 + 5x \\ 8 + 40x + 5 = 13 + 5x \\ 40x + 13 = 13 + 5x \\ 40x - 5x = 13 - 13 \\ 35x = 0 \\ x = 35 \times 0 \\ x = 0$

7 0

3 years ago

Three fewer than one fourth of x is 12

Wittaler [7]

The answer is 1/4 x - 3 = 12

6 0

4 years ago

Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.Tiles

Ugo [173]

By definition, the average rate of change is given by:

$AVR = \frac{f(x2)-f(x1)}{x2-x1}$

We evaluate each of the functions in the given interval.

We have then:

For f (x) = x ^ 2 + 3x:

Evaluating for x = -2:

$f (-2) = (-2) ^ 2 + 3 (-2)\\f (-2) = 4 - 6\\f (-2) = - 2$

Evaluating for x = 3:

$f (3) = (3) ^ 2 + 3 (3)\\f (3) = 9 + 9\\f (3) = 18$

Then, the AVR is:

$AVR = \frac{18-(-2)}{3-(-2)}$

$AVR = \frac{18+2}{3+2}$

$AVR = \frac{20}{5}$

$AVR = 4$

For f (x) = 3x - 8:

Evaluating for x =4:

$f (4) = 3 (4) - 8\\f (4) = 12 - 8\\f (4) = 4$

Evaluating for x = 5:

$f (5) = 3 (5) - 8\\f (5) = 15 - 8\\f (5) = 7$

Then, the AVR is:

$AVR = \frac{7-4}{5-4}$

$AVR = \frac{3}{1}$

$AVR = 3$

For f (x) = x ^ 2 - 2x:

Evaluating for x = -3:

$f (-3) = (-3) ^ 2 - 2 (-3)\\f (-3) = 9 + 6\\f (-3) = 15$

Evaluating for x = 4:

$f (4) = (4) ^ 2 - 2 (4)\\f (4) = 16 - 8\\f (4) = 8$

Then, the AVR is:

$AVR = \frac{8-15}{4-(-3)}$

$AVR = \frac{-7}{4+3}$

$AVR = \frac{-7}{7}$

$AVR = -1$

For f (x) = x ^ 2 - 5:

Evaluating for x = -1:

$f (-1) = (-1) ^ 2 - 5\\f (-1) = 1 - 5\\f (-1) = - 4$

Evaluating for x = 1:

$f (1) = (1) ^ 2 - 5\\f (1) = 1 - 5\\f (1) = - 4$

Then, the AVR is:

$AVR = \frac{-4-(-4)}{1-(-1)}$

$AVR = \frac{-4+4}{1+1}$

$AVR = \frac{0}{2}$

$AVR = 0$

Answer:

from the greatest to the least value based on the average rate of change in the specified interval:

f(x) = x^2 + 3x interval: [-2, 3]

f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]

4 0

3 years ago

You run at a rate of 4 mph and your friend runs at a rate of 3.5 mph Your friend starts running 10 minutes before you, and you b

irga5000 [103]

Answer:

A)

i) Your speed: 0.067 miles per minute

ii) Your friend's speed: 0.058 miles per minute

B) You will not catch up with your friend by the time your friend stops running.

Explanation:

A. Find your speed and your friend's speed in miles per minute

i) Your speed:

You know your speed is 4 mph, which is 4 miles per hour. To find you spedd in miles per minute you must multiply by the factor that relates hours and minutes:

1 hour = 60 min

1 = 1 hour / 60 min

4 miles / hour × 1 hour / 60 min = 4 miles / 60 min ≈ 0.0667 miles/min

(note that the unit hour is in the denominator of the first fraction and in the numerator of the second fraction, so they cancel each other, and the final units are miles / min).

ii) Your friend's speed

3.5 miles / hour × 1 hour / 60 min = 3.5 miles / 60 min ≈ 0.0583 miles / min.

(again, the unit hour is in the denominator of the first fraction and in the numerator of the second fraction, so they cancel each other, and the final units are miles / min).

B. Will you catch up with your friend by the time your friend stops running?

To find whether you catch up your friend you can use a system of equation that relates the times and the distances run by each one.

i) The distance run by you is equal to your time times your speed:

d₁ = 0.0667 miles / min × t₁

ii) The distance run by your friend is equal to his time times his speed.

d₂ = 0.0583 miles / min × t₂

Since you friend started running 10 minutes before you, his time is 10 minutes more than yours, i.e. t₂ = 10 min + t₁.

Now you can write the second equation as:

d₂ = 0.0583 miles / min × (10 min + t₁).

And since you want to determine whether you catch up him/her by the time he/she stops running, you can determine the time when both distances would be equal:

d₁ = d₂

0.0667 miles / min × t₁ = 0.0583 miles / min × (10 min + t₁)

To solve the equation I will ignore the units (normal procedure to solve equations):

0.0667 t₁ = 0.0583 (10 + t₁)

0.0667 t₁ = 0.583 + 0.0583t₁

0.0667 t₁ - 0.0583 t₁ = 0.583

0.0084 t₁ = 0.583

t₁ = 0.583 / 0.0084 ≈ 69 min

That is the time you need to catch up your friend, 69 min. Since your friend started 10 min before you and will stop after half an hour (30 min), you will not catch up with your friend before the time your friend stops running.

5 0

4 years ago