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3 years ago

!!!!!!PLEASE HELP WILL GIVE BRAINLIEST!!!!!!!

Mathematics

1 answer:

lawyer [7]3 years ago

3 0

<h2>B is the correct answer</h2><h3>Let me know if I am wrong.</h3>

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Suppose it is known that the distribution of purchase amounts by customers entering a popular retail store is approximately norm

Ket [755]

Answer:

69.15% probability that a randomly selected customer spends less than $105 at this store

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 100, \sigma = 10$

What is the probability that a randomly selected customer spends less than $105 at this store?

This is the pvalue of Z when X = 105. So

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

$Z = \frac{105 - 100}{10}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

69.15% probability that a randomly selected customer spends less than $105 at this store

7 0

3 years ago

Read 2 more answers

For which values of the parameter B the following system admits a unique solution [ 1 b 1-b 2 2 0 2–2B 4 0] [x y z] = [1 2 0]

photoshop1234 [79]

Answer: β ≠ ±1

Step-by-step explanation: For a system of equations to have an unique solution, its determinant must be different from 0: det |A| ≠ 0. So,

det $\left[\begin{array}{ccc}1&\beta&1-\beta\\2&2&0\\2-2\beta&4&0\end{array}\right]$ ≠ 0

Determinant of a 3x3 matrix is calculated by:

det $\left[\begin{array}{ccc}1&\beta&1-\beta\\2&2&0\\2-2\beta&4&0\end{array}\right]\left[\begin{array}{ccc}1&\beta\\2&2\\2-2\beta&4\end{array}\right]$

$8(1-\beta)-[2(2-2\beta)(1-\beta)]$

$8-8\beta-4+8\beta-4\beta^{2}$

$-4\beta^{2}+4\neq 0$

$\beta^{2}\neq 1$

$\beta \neq \sqrt{1}$

β ≠ ±1

For the system to have only one solution, β ≠ 1 or β ≠ -1.

7 0

3 years ago

Two number cubes are rolled.

yaroslaw [1]

Answer:

1/4

Step-by-step explanation:

Assuming the number cube is a six-faced die, you have

1 2 3 4 5 6

three odd numbers, and three even numbers. Therefore, the chance of it landing on an odd number or even number is 3/6, which equals 1/2. That means you have a 50% chance to get an odd number, or an even number.

So, you have two die. Both have a 50/50 chance of getting an even or odd number. So what's the chance of one landing on an odd number, and the other landing on an even number?