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Archy [21]
3 years ago
10

On average students attend school 170 of the 180 days required in a year. About what part of the required days do they attend sc

hool?
Mathematics
2 answers:
Llana [10]3 years ago
3 0
94% of the required days, if you wanted percentages
Delicious77 [7]3 years ago
3 0
94% of the required days, if you wanted percentagesBRAINLIEST???

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find the total cost of a $55.00 dinner bill including 7% tax and 18% tip.how much change will you get back from $100.00
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The total cost of the dinner with a pre-tax tip would be $65.78. If you paid this with a $100 bill, you would get $34.22 back.

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3 years ago
Whats the answer then
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Wheres the question??
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Step-by-step explanation:

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For what value of a does (1/7)^3a+3=343^a-1
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2 years ago
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
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