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OverLord2011 [107]
3 years ago
5

Mrs. Santos baked 25 cookies. 3 fifths of the cookies were oatmeal. How many oatmeal cookies did Mrs. Santos bake?

Mathematics
1 answer:
grigory [225]3 years ago
7 0

Answer: 15 oatmeal cookies.

Step-by-step explanation:

There were 25 cookies baked.

Out of these, 3/5 are oatmeal cookies.

The amount of actual oatmeal cookies is:

= fraction of oatmeal cookies * total number of cookies

= 3/5 * 25

= 75 / 5

= 15 oatmeal cookies.

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A waiter received $3 tip. If the check totaled $25, what percent tip was received?
EleoNora [17]

Answer: 12%

Step-by-step explanation: Percent easily found with 100. 25*4=100. 3*4=12

12/100=12%

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F(x) = –2x2 + 17<br> Find f(10)
nikdorinn [45]

Answer:

If its -2x^2+17 the answer would be -183.

Step-by-step explanation:

-2(100)+17

-200+17

= -183

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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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