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zavuch27 [327]
3 years ago
13

Troy spends $9.75 on chicken feed

Mathematics
1 answer:
Oxana [17]3 years ago
4 0

Answer:

A

Step-by-step explanation:

Total eggs in lay in one week = 48 * 7 = 336

total eggs sold = 336 - 12 = 324

Total dozen = 324 : 12 = 27

Total profit = 27 * 4 = $108

Profit without chicken feed = 108 - 9.75 = $98.25

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Which pair of angles is complementary?
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Answer:

2 angles are shown one angle is 56 degrees and the other is 34 degrees

Step-by-step exp

complementary angle=90 degree

56+34=90

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3 years ago
Hey can you please help me posted pictureof question
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There are two 5's because the 5 section takes up the amount of 2 sections.

There is a total of 8 sections because sections 2 and 5 take up the amount of 2 sections

5's to total = 2/8 = 1/4

Your answer is A. 1/4

Hope this helps :)
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What is the solution to the equation 1.6m − 4.8 = −1.6m?
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2 years ago
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Yolanda wants to pour 91.48 grams of salt into a container so far she has poured 72.2 grams how much more salt should Yolanda po
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Answer:

19.28

Step-by-step explanation:

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2 years ago
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In 2008, data from the Center for Disease Control revealed that 28.5% of all male teenagers, aged 18-19 and attending U.S. colle
m_a_m_a [10]

Answer:

a) H0 : u = 28.5%

H1 : u < 28.5%

b) critical value = - 1.645

c) test statistic Z= - 1.41

d) Fail to reject H0

e) There is not enough evidence to support the professor's claim.

Step-by-step explanation:

Given:

P = 28.5% ≈ 0.285

X = 210

n = 800

p' = \frac{X}{n} = \frac{210}{800} = 0.2625

Level of significance = 0.05

a) The null and alternative hypotheses are:

H0 : u = 28.5%

H1 : u < 28.5%

b) Given a 0.05 significance level.

This is a left tailed test.

The critical value =

-Z_0.05 = -1.645

The critical value = -1.645

c) Calculating the test statistic, we have:

Z = \frac{p' - P}{\sqrt{\frac{P(1-P)}{n}}}

Z = \frac{0.2625 - 0.285}{\sqrt{\frac{28.5(1-28.5)}{800}}}

Z = -1.41

d) Decision:

We fail to reject null hypothesis H0, since Z = -1.41 is not in the rejection region, <1.645

e) There is not enough evidence to support the professor's claim that the proportion of obese male teenagers decreased.

5 0
3 years ago
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