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1 year ago

Find the sum of the average monthly rainfall. this is on i ready okay so please help.

1 answer:

6 0

Based on the average monthly rainfall amounts as shown in the graph, the sum of the average monthly rainfall 11/12 inches.

<h3>What is the sum of the average monthly rainfall?</h3><h3 />

The average can be found by adding up the monthly rainfall and dividing by 12 months.

Solving gives:

= (1/8 + 1/8 + 1/8 + 2/8+ 2/8 + 5/8 + 7/8 + 9/8 + 13/8 + 15/8 + 16/8 + 16/8) / 12

= (88/8) / 12

= 11/12 inches

Find out more on average rainfall at brainly.com/question/23871385

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F(x) = -x² + 4x - 1 find the value of f(x) if x = 0

ollegr [7]

Answer:

-1

Step-by-step explanation:

Given,

f(x) = -x² + 4x - 1 and x = 0

Putting the value of x we get,

f(0)= $-0^{2}$ + (4 X 0) - 1

= -0+0-1

=-1

∴f(x)= -1

So, the required value is -1

8 0

1 year ago

What is the intersection of the following sets? G = {3, 7, 8, 9} H = {2, 5, 7, 8}

weeeeeb [17]

"The intersection (∩) of a pair of sets (G and H) is a third set (I) composed by the elements that belong, at the same time, to both given sets."

According to this definition:

Given the sets:

G = {3, 7, 8, 9}

H = {2, 5, 7, 8}

The Intersection is:

G ∩ H = I = {7, 8}

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5 0

2 years ago

Read 2 more answers

Matteo is adjusting a satellite because he finds it is not focusing the incoming radio waves perfectly. The shape of his satelli

natka813 [3]

Answer: (1.5, 0)

Step-by-step explanation:

Given : The shape of his satellite can be modeled by $y^2=6x$ where x and y are modeled in inches.

Now,the given equation $y^2=6x$ is a equation of parabola.

Here, coefficient of x is positive, hence the parabola opens rightwards.

On comparing this equation with standard equation $x^2=4mx$ , we get

$4m=6\\\Rightarrow\ m=\frac{6}{4}=1.5$

In standard equation, coordinates of focus=(m,0)

Thus for given equation coordinates of focus=(1.5,0)

4 0

2 years ago

Read 2 more answers

PLEASE HELP. I WILL GIVE BRAINLIEST!<br> I'M HAVING SOME TROUBLE.

nalin [4]

Answer:

A.) The coordinates of M' are (-2/4, -4/5)

The coordinates of N' are (1/5, 1)

B.) The scale factor of the dialation is 3

7 0

2 years ago

Suppose that scores on an aptitude test are normally distributed with a mean of 100 and a standard deviation of 4.6. Scores on a

DerKrebs [107]

Answer:

<u>The correct answer is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.</u>

Step-by-step explanation:

1. Let's check all the information given to us to answer the question correctly:

Mean of the scores on the aptitude test = 100

Standard deviation of the aptitude test = 4.6

Felix's score on the aptitude test = 106

Mean of the scores on the knowledge test = 70

Standard deviation of the knowledge test = 2.4

Felix's score on the knowledge test = 75

2. Which statement best describes Felix's scores on the two tests comparatively?

Let's recall that z-score in a normal distribution, positive or negative, is the number of times of the standard deviation a certain element is from the mean. If the element is below the mean, then the z-score is negative and if it's above the mean, then the z-score is positive.

Therefore, a score of 106 on the aptitude test, will have the following z-score:

106 - 100 = 6 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

6/4.6 = 1.30

A score of 75 on the knowledge test, will have the following z-score:

75 - 70 = 5 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

5/2.4 = 2.08

The z-scores of Felix were 1.30 on the aptitude test and 2.08 on the knowledge test. He performed better on the aptitude test because 2.08 > 1.30, so the correct statement that best describes Felix's scores on the two tests comparatively is<u> C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.</u>

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2 years ago