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enot [183]
3 years ago
6

Weekly demand for a product is normally distributed with a mean and standard deviation of 3,211 and 484 units, respectively. Wha

t is the probability that weekly demand exceeds 4,044
Mathematics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

The probability it is greater than or equal to 0.95 and less than or equal to 0.96

Step-by-step explanation:

In order to calculate the probability that weekly demand exceeds 4,044 we would have to calculate the z value as follows:

Z= (X-mean)/standard deviation

Z= (4044 - 3211)/484

Z=1.721

So, looking for standard normal distribution table 1.7 from row, we get values from 0.9554 till 0.9633, for column 0.02 it is 0.9573

Therefore, the probability it is greater than or equal to 0.95 and less than or equal to 0.96

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earnstyle [38]

Answer:

7m^4-52m^3-28m^2+67m+42

Step-by-step explanation:

(7m^2-3m-7)(m^2-7m-6)

7m^4-3m^3-7m^2-49m^3+21m^2+49m-42m^2+18m+42

7m^4-3m^3-49m^3-7m^2+21m^2-42m^2+49m+18m+42

7m^4-52m^3-28m^2+67m+42

4 0
3 years ago
A teacher is making identical packets using 92 crayons and 23 sheets of paper. What is the greatest number of packets she can ma
Natali [406]
92 crayons
23sheets of paper
92÷23=4
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3 years ago
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Paul owns a mobile wood-fired pizza oven operation. A couple of his clients complained about his dough at a recent catering, so
Makovka662 [10]

Answer: B. Do not reject H0, we cannot conclude the proportion of customer complaints is more for the old dough

Step-by-step explanation:

This is a test of 2 population proportions. Let 1 and 2 be the subscript for the old and the new dough players. The population proportions of customer complaints with the old and new dough would be p1 and p2

P1 - P2 = difference in the proportion of customer complaints with the old and new dough.

The null hypothesis is

H0 : p1 ≥ p2

pm - pw ≥ 0

The alternative hypothesis is

Ha : p1 < p2

p1 - p2 < 0

it is a left tailed test

Sample proportion = x/n

Where

x represents number of success(number of complaints)

n represents number of samples

For old dough

x1 = 6

n1 = 385

P1 = 6/385 = 0.016

For new dough,

x2 = 16

n2 = 340

P2 = 16/340 = 0.047

The pooled proportion, pc is

pc = (x1 + x2)/(n1 + n2)

pc = (6 + 16)/(385 + 340) = 0.03

1 - pc = 1 - 0.03 = 0.97

z = (Pm - Pw)/√pc(1 - pc)(1/nm + 1/nw)

z = (0.016 - 0.047)/√(0.03)(0.97)(1/385 + 1/340) = - 0.031/√0.00553857907

z = - 0.42

Since it is a left tailed test, we would find the p value for the area to the left of the z score. From the normal distribution table,

p value = 0.337

For a 95% confidence level, the significant level, alpha is

1 - 0.95 = 0.05

Since 0.05 < 0.337, we would accept the null hypothesis

Therefore, Do not reject H0, we cannot conclude the proportion of customer complaints is more for the old dough

7 0
3 years ago
Find the slope to Y = 5/2x-3
son4ous [18]

Answer:

slope = 5/2

Step-by-step explanation:

This equation is written in slope intercept form

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y = 5/2x -3

The slope is 5/2 and the y intercept is -3

5 0
3 years ago
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Can someone explain to me how to do question number 6 please?
slavikrds [6]
Hope this helps but try to do process of elimination.
7 0
3 years ago
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