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2 years ago

#1) a normal distribution has a mean of 101 and a standard deviation of 4.2. what is the z-score of 105?

Mathematics

1 answer:

nika2105 [10]2 years ago

5 0

Step-by-step explanation:

z = (specific score - mean) / standard deviation

in our case

z = (105 - 101)/4.2 = 4/4.2 = 0.952380952... ≈ 0.95

as z-tables usually round the z-scores to hundredths.

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The difference is 1,302 feet.

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3 years ago

A park has a grinkgo tree a dogwood tree and 2 blue spruce trees the blue spruce trees are 8 years old the grinkgo tree is 2 yea

enot [183]

Answer:

The age of the grinkgo tree is 10 years old and the age of the dogwood tree is 4 years old

Step-by-step explanation:

Let

x -----> the age of the grinkgo tree

y -----> the age of the dogwood tree

z -----> the age of the blue spruce tree

we know that

$x=3y-2$ ------> equation A

$z=8$ ----> equation B

$x=\frac{1}{2}(y+2z)$

substitute the value of z

$x=\frac{1}{2}(y+2(8))$

$x=\frac{1}{2}(y+16)$ -----> equation C

equate equation A and equation C and solve for y

$3y-2=\frac{1}{2}(y+16)$

$6y-4=(y+16)$

$6y-y=16+4$

$5y=20$

$y=4$

Find the value of x (equation A or equation C)

$x=3(4)-2=10$

therefore

The age of the grinkgo tree is 10 years old and the age of the dogwood tree is 4 years old

7 0

4 years ago

Find r for the geometric series with s5 = 484, a1 = 4, a5 = 324

sleet_krkn [62]

A5 / a1 = r^4 = 324/4

So r^4 = 81

r = +/- 3

If r = 3 :-

S5 = 4 * (3^5-1) / (3-1) = 484

so r = 3 Answer

4 0

3 years ago

The measure of an angle is 22° greater than it’s complement. Find the angle

mamaluj [8]

Answer:

Step-by-step explanation:

Let the angle be " x ".

Complement of the angle = ( x - 22 )

To find : x = ?

x + x - 22 = 90

2x - 22 = 90

2x = 90 + 22

2x = 112

x = 112 / 2

x = 56

Therefore, the angle is 56.

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3 years ago

Johnson middle school had 240 students last year. This year. There are 192 students. Find the percent of change. State whether t

Rina8888 [55]

Answer:

Step-by-step explanation:

192 ÷ 420 = 0.457

0.457 x 100 = 45.7

But there can't be a percentage increase or decrease because there is no percentage to compare it to, there should be information for the change in students of the last year

7 0

4 years ago