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EastWind [94]
3 years ago
6

(02.02 MC) Use the graph to fill in the blank with the correct number.

Mathematics
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

The answer is "2"

Step-by-step explanation:

When we check the points where the x is -2

by calculating the "y-value":

It is 2, right?  

it implies that f(-2)=2

that's why the final answer is "2"

You might be interested in
Which of the following rational numbers is equal to that? please help quickly
zloy xaker [14]

Answer:

31/9. you have to count by nines, get three whole numbers which are not fractions, and you are left with 4. and a whole 3 plus 0.4 = 3.4 hope that helps homie

3 0
3 years ago
In the United States, 7% of all registered voters belong to the Green party. A random sample of 50 registered voters is taken. U
Maksim231197 [3]

Answer:

The expected value of the sample proportion is of 0.07.

1. P(p < .02) = 0.0823

2. P(p > .15) = 0.0132

3. P(.05 < p < .09) = 0.4176

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

7% of all registered voters belong to the Green party. 50 voters:

This means that p = 0.07, n = 50

So, for the normal distribution:

\mu = 0.07, s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.07*0.93}{50}} = 0.036

The expected value of the sample proportion is of 0.07.

1. Determine P(p < .02).

This is the pvalue of Z when X = 0.02. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.02 - 0.07}{0.036}

Z = -1.39

Z = -1.39 has a pvalue of 0.0823

So

P(p < .02) = 0.0823

2. Determine P(p > .15).

This is 1 subtracted by the pvalue of Z when X = 0.15. So

Z = \frac{X - \mu}{s}

Z = \frac{0.15 - 0.07}{0.036}

Z = 2.22

Z = 2.22 has a pvalue of 0.9868

1 - 0.9868 = 0.0132

So

P(p > .15) = 0.0132

3. Determine P(.05 < p < .09).

This is the pvalue of Z when X = 0.09 subtracted by the pvalue of Z when X = 0.05. So

X = 0.09

Z = \frac{X - \mu}{s}

Z = \frac{0.09 - 0.07}{0.036}

Z = 0.55

Z = 0.55 has a pvalue of 0.7088

X = 0.05

Z = \frac{X - \mu}{s}

Z = \frac{0.05 - 0.07}{0.036}

Z = -0.55

Z = -0.55 has a pvalue of 0.2912

0.7088 - 0.2912 = 0.4176. So

P(.05 < p < .09) = 0.4176

3 0
3 years ago
Four times a number is three times the difference of that number and one
andrey2020 [161]

Answer: the number is -3

Step-by-step explanation:

Four times a number is three times the difference of that number and one

Let the number be x. Four times the number means 4×x = 4x

three times the difference of that number and one. This means

Difference of the number and one is (x-1). Three times the difference of the number and one would be 3(x-)

Four times a number = three times the difference of that number and one. Therefore,

4x = 3(x-1)

Opening the bracket

4x = 3x -3

4x-3x = -3

x =-3

Check

4×-3 =3(-3-1)

-12 = -3×4

-12 = -12

So the number is -3

3 0
3 years ago
PLEASE HELP!
andreyandreev [35.5K]

Answer:

2,4,1

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
a. it is a multiple of 3,b. is greater then 100 ,c. it is not even, d.all digits are odd. whats the number?
uranmaximum [27]

Answer:

333.

Step-by-step explanation:

3•111=333

333>100

All the digits are odd...


4 0
3 years ago
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