Which is the graph of 4x

What would A,B,C,D.E be?

Anna11 [10]

Answer:

Between 0-180

Step-by-step explanation:

Cause it is facts first

4 0

3 years ago

Read 2 more answers

Since 1900, the magnitude of earthquakes that measure 0.1 or higher on the Richter Scale in CA are distributed normally with a m

Gwar [14]

Answer:

a) 3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1

b) 1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1

c) 73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1

d) 99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

e) 6.0735

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using 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:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 6.2, \sigma = 0.5$

a.) What is the probability that a randomly selected earthquake in CA has a magnitude greater than 7.1?

This is 1 subtracted by the pvalue of Z when X = 7.1. So

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

$Z = \frac{7.1 - 6.2}{0.5}$

$Z = 1.8$

$Z = 1.8$ has a pvalue of 0.9641

1 - 0.9641 = 0.0359

3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1

b.) What is the probability that a randomly selected earthquake in CA has a magnitude less than 5.1?

This is the pvalue of Z when X = 5.1. So

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

$Z = \frac{5.1 - 6.2}{0.5}$

$Z = -2.2$

$Z = -2.2$ has a pvalue of 0.0139

1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1

c.) What is the probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1?

Now $n = 10, s = \frac{0.5}{\sqrt{10}} = 0.1581$

This is 1 subtracted by the pvalue of when X = 6.1. So

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

By the Central Limit Theorem

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

$Z = \frac{6.1 - 6.2}{0.1581}$

$Z = -0.63$

$Z = -0.63$ has a pvalue of 0.2643

1 - 0.2643 = 0.7357

73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1

d.) What is the probability that a ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

This is the pvalue of Z when X = 7.22 subtracted by the pvalue of Z when X = 5.7. So

X = 7.22

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

$Z = \frac{7.22 - 6.2}{0.1581}$

$Z = 6.45$

$Z = 6.45$ has a pvalue of 1

X = 5.7

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

$Z = \frac{5.7 - 6.2}{0.1581}$

$Z = -3.16$

$Z = -3.16$ has a pvalue of 0.0008

1 - 0.0008 = 0.9992

99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22

e.) Determine the 40th percentile of the magnitude of earthquakes in CA.

This is X when Z has a pvalue of 0.4. So it is X when Z = -0.253.

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

$-0.253 = \frac{X - 6.2}{0.5}$

$X - 6.2 = -0.253*0.5$

$X = 6.0735$

6 0

3 years ago

The brand xyz dog food costs $36 for 15 pounds of dog food. What is the cost per pound?

HACTEHA [7]

Answer:

Step-by-step explanation:

36/$15 = $2.04 per pound

3 0

3 years ago

Read 2 more answers

Suppose I pick a jelly bean at random from a box containing two red and ten blue ones. I record the color and put the jelly bean

Lubov Fominskaja [6]

Answer:

The probability of getting a red jelly bean both of the times is 0.028.

Step-by-step explanation:

First we have that in the box are 12 jelly beans having 2 red and 10 blue. The probability of getting one of the red jelly beans are the sum of the probability of the 2 red jelly beans, that is:

$ProbabilityRed=\frac{1}{12} +\frac{1}{12}=\frac{2}{12}$

Now we can calculate the probability of getting the red jelly bean twice, this will be the product of the probability of getting each one, is the same probability for the second chance because we are getting the jelly bean back in the box, this is:

$ProbabilityBothRed=\frac{2}{12} *\frac{2}{12}=\frac{4}{144} =\frac{1}{36} =0.028$

Now we have that the probability of getting a red jelly bean both of the times is 0.028.

6 0

4 years ago

Which of the following graphs represents a function choose all that apply

melisa1 [442]

Answer:

where the graphs?

Step-by-step explanation:

4 0

3 years ago

Which is the graph of 4x – 3y = 12?