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3 years ago

SOMEONE IN THIS APp can you please help please please ASAP I posted the question 3 times come on y’all help please

Mathematics

1 answer:

Shalnov [3]3 years ago

3 0

Angle 1 - 72 degrees

Angle 2 - 36 degrees

Angle 3 - 18 degrees

Angle 4 - 144 degrees

Hope this helped!

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Last June only 0.17 inches of rain fell all month. What is the difference between the average rainfall and the actual rainfall f

spin [16.1K]

Answer:

Part 1) 0.50 inches

Part 2) 1.14 inches

Part 3)

a) 1.25 inches

b) -0.32 inches

Step-by-step explanation:

The complete question in the attached figure

Part 1) Last June only 0.17 inches of rain fell all month. What is the difference between the average rainfall and the actual rainfall for the last June

Let

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

The difference between the average rainfall and the actual rainfall for the last June is given by the expression

$(x-y)$

we have

$x=0.67\ in\\y=0.17\ in$

substitute

$0.67-0.17=0.50\ in$

Part 2) The departure from the average rainfall last July was -0.36 inches. How much rain fell last July?

we know that

The departure from the average is the difference between the actual amount of rain and the average amount of rain for a given month

Let

D -----> the departure from the average

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

$D=y-x$

we have

$D=-0.36\ in\\x=1.5\ in$

substitute the given values

$-0.36=y-1.5$

solve for y

$y=1.5-0.36\\y=1.14\ in$

Part 3)

Part a) How much rain would have to fall in August so that the total amount of rain equals the average rainfall for these three months?

Part b) What would the departure from the average be in August in that

situation?

Part a) How much rain would have to fall in August so that the total amount of rain equals the average rainfall for these three months?

Find the average rainfall for these three months

Adds the average rainfall and divide by 3

$(0.67+1.5+1.57)/3=1.25\ in$

Part b) What would the departure from the average be in August in that

situation?

we know that

The departure from the average is the difference between the actual amount of rain and the average amount of rain for a given month

Let

D -----> the departure from the average

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

$D=y-x$

we have

$x=1.57\ in\\y=1.25\ in$

substitute the given values

$D=1.25-1.57=-0.32\ in$

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4 years ago

Quation that show an estimate of 7,203+299

Dmitry [639]

The two numbers are very near 7200 and 300. The sum of those is a good estimate of your sum:

... 7200 + 300 = 7500

_____

If you're taught methods of addition that include "make a ten" or "make a hundred", you know you can "borrow" a 1 from 7203 (making it 7202) and loan that 1 to 299 to make it 300. Now, you know your sum is exactly 7202+300 = 7502, with no estimation necessary.

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3 years ago

Read 2 more answers

How many units is the perimeter of a rectangle with vertices located at (–2,2), (–2,–2), (4,–2), and (4,2

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Answer:

Step-by-step explanation:

The perimeter is the sum of the side lengths:

4 + 6 + 4 + 6 = 20 . . . . units

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4 years ago

A student faculty government committee of 4 people is to be formed from 20 student volunteers and 5 faculty volunteers. Find the

Arte-miy333 [17]

A.) Probability of selecting two students and two faculty members from 20 students and 5 faculty volunteers = no of ways of selecting two students and two faculty members from 20 students and 5 faculty volunteers divided by the number of ways of selecting 4 people from 25 people = (20C2 x 5C2) / 25C5 = 1900 / 53130 = 0.036 = 3.58%

b.) Probability of selecting three students and one faculty members from 20 students and 5 faculty volunteers = no of ways of selecting three students and one faculty members from 20 students and 5 faculty volunteers divided by the number of ways of selecting 4 people from 25 people = (20C3 x 5C1) / 25C5 = 5700 / 53130 = 0.107 = 10.7%

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3 years ago

It takes Jack 11 hours to pick 40 bushels of apples alone. Jill can pick 40 bushels of apples in 9 hours. How long will it take

elena55 [62]

Answer:

5 hours

Step-by-step explanation:

Jack 40/11 = 3.64 bushels per hour

Jill 40/9 = 4.44 bushels per hour

together they pick 3.64 + 4.44 = 8.08 bushels per hour

40/8.08 = 4.95 hours

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3 years ago