All categories

2 years ago

Please help me and no links this is due in 10 mins ill give you the crown

Mathematics

1 answer:

FinnZ [79.3K]2 years ago

7 0

Answer:

angle Q is 68°

angle R is 112°

You might be interested in

Elizabeth earned a grade of 85% on her multiple-choice science final that had a total of 160 problems. How many problems on the

Naily [24]

Answer:

135

Step-by-step explanation: Multiply 160 by 0.85(for 85%) and you get 135

7 0

3 years ago

1 point

kondor19780726 [428]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

A number was subtracted from 15. After which, that result was multiplied by 5. This result was then divided by 5 for a result of

likoan [24]

Answer:

Step-by-step explanation:

15-14 = 1

1*5=5

5/1=1

6 0

2 years ago

Factor 64v^3 + 192v^2 -56v - 168

Kaylis [27]

Hello from MrBillDoesMath!

Answer:

8(v+3) ( -1/2 (sqrt(14) - 4 v) (4 v + sqrt(14)) )

Discussion:

Given

64v^3 + 192v^2 - 56 v - 168

Factor 64v^2 from the first two terms. Factor 56 from the last two terms:

64v^2(v+3) - 56(v + 3) => factor (v+3) from both terms

(v+3) (64v^2 - 56) => factor 8 from both terms in the right ()

8(v+3)(8v^2-7) => factor 8y^2-7

8(v+3) ( -1/2 (sqrt(14) - 4 v) (4 v + sqrt(14)) )

Thank you,

MrB

4 0

3 years ago

Normal Distribution. Cherry trees in a certain orchard have heights that are normally distributed with mu = 112 inches and sigma

Lubov Fominskaja [6]

Answer:

The probability that a randomly chosen tree is greater than 140 inches is 0.0228.

Step-by-step explanation:

Given : Cherry trees in a certain orchard have heights that are normally distributed with $\mu = 112$ inches and $\sigma = 14$ inches.

To find : What is the probability that a randomly chosen tree is greater than 140 inches?

Solution :

Mean - $\mu = 112$ inches

Standard deviation - $\sigma = 14$ inches

The z-score formula is given by, $Z=\frac{x-\mu}{\sigma}$

Now,

$P(X>140)=P(\frac{x-\mu}{\sigma}>\frac{140-\mu}{\sigma})$

$P(X>140)=P(Z>\frac{140-112}{14})$

$P(X>140)=P(Z>\frac{28}{14})$

$P(X>140)=P(Z>2)$

$P(X>140)=1-P(Z$

The Z-score value we get is from the Z-table,

$P(X>140)=1-0.9772$

$P(X>140)=0.0228$

Therefore, the probability that a randomly chosen tree is greater than 140 inches is 0.0228.

5 0

3 years ago