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3 years ago

Can someone help me fill out this chart

Mathematics

1 answer:

kotegsom [21]3 years ago

6 0

Answer:

is there a y part of this equation?

Step-by-step explanation:

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4- A manufacturing process produces items whose weights are normally distributed. It is known that 22.57% of all the items produ

galben [10]

Answer:

$\\ \mu = 118\;grams\;and\;\sigma=30\;grams$

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find $\\ \mu\;and\;\sigma$ .

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

$\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}$

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

$\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}$

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

<h3>Solving a system of equations for values of the mean and standard deviation</h3>

Having equations (1) and (2), we can form a system of two equations and two unknowns values:

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

Rearranging these two equations:

$\\ -0.6*\sigma = 100-\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

To solve this system of equations, we can multiply (1) by -1, and them sum the two resulting equation:

$\\ 0.6*\sigma = -100+\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

Summing both equations, we obtain the following equation:

$\\ 3.0*\sigma = 90$

Then

$\\ \sigma = \frac{90}{3.0} = 30$

To find the value of the mean, we need to substitute the value obtained for the standard deviation in equation (2):

$\\ 2.4*30 = 190-\mu$ (2)

$\\ 2.4*30 - 190 = -\mu$

$\\ -2.4*30 + 190 = \mu$

$\\ \mu = 118$

7 0

3 years ago

Determine whether the given measures can be the lengths of the sides of a triangle write yes or no

ohaa [14]

NO.,the given measures can not be the lengths of the sides of a triangle

Step-by-step explanation

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

so, Find the range for the measure of the third side of a triangle given the measures of two sides.

here given measures are 2,2,6

2+2 = 4 which is less than the third side 6

= 4 < 6

This not at all a triangle.

Hence, the given measures can not be the lengths of the sides of a triangle

3 0

4 years ago

I need help on the top one 25 pts

svet-max [94.6K]

The answer to the question

8 0

3 years ago

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper

Korvikt [17]

Answer:

90.67% probability that John finds less than 7 golden sheets of paper

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it contains a golden sheet of paper, or it does not. The probability of a container containing a golden sheet of paper is independent of other containers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

At Munder Difflin Paper Company, the manager Mitchell Short randomly places golden sheets of paper inside of 30% of their paper containers.

This means that $p = 0.3$

14 of these containers of paper.

This means that $n = 14$

What is the probability that John finds less than 7 golden sheets of paper?

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{14,0}.(0.3)^{0}.(0.7)^{14} = 0.0068$

$P(X = 1) = C_{14,1}.(0.3)^{1}.(0.7)^{13} = 0.0407$

$P(X = 2) = C_{14,2}.(0.3)^{2}.(0.7)^{12} = 0.1134$

$P(X = 3) = C_{14,3}.(0.3)^{3}.(0.7)^{11} = 0.1943$

$P(X = 4) = C_{14,4}.(0.3)^{4}.(0.7)^{10} = 0.2290$

$P(X = 5) = C_{14,5}.(0.3)^{5}.(0.7)^{9} = 0.1963$

$P(X = 6) = C_{14,6}.(0.3)^{6}.(0.7)^{8} = 0.1262$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0068 + 0.0407 + 0.1134 + 0.1943 + 0.2290 + 0.1963 + 0.1262 = 0.9067$

90.67% probability that John finds less than 7 golden sheets of paper

7 0

3 years ago

PLEASE HELPPPPP MEEE♥️♥️♥️♥️

Masteriza [31]

Answer:

$(.2246 \\ 5513$

5 0

3 years ago