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3 years ago

Tamara feeds her cat 1/8cup canned food each day and the

2 answers:

8 0

Do not copy Faroquem9’s work I just used it to answer your question and also I just used it as a example and wrote it on my own way so don’t copy it use it as a example and write it on your own way or answer it on your own way

AND THIS IS NOT MY WORK IT IS Faroquem9’s work..

Dmitry_Shevchenko [17]3 years ago

3 0

Answer:

well first u have to pray. ex: "oh the holy spirit pls forgive my sins...." "I love eating cornbread" ❤❤❤❤❤❤❤❤

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What is the least common multiple of 4 and 10?<br> 02<br> 0 4<br> 20<br> 40

viktelen [127]

Answer: 4 I think

Step-by-step explanation:

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2 years ago

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Write the equation of a line with a slope of -2 and a y-intercept of 5:

saveliy_v [14]

Answer: y = -2x +5 , so C (you probably just forgot to put the x)

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3 years ago

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First discovered more than 30 years ago, Lina’s sunbird, a four-and-a-half-inch animal found in the Philippines and that resembl

Arlecino [84]

Answer: the correct one is C) found in the Philippines and resembling

Step-by-step explanation:

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3 years ago

A 5 hour conference booking is required for an office party for 120 people which includes a buffet dinner. Hotel 5 has recently

SpyIntel [72]

Answer: I think that the cost now from hotel 5 would be $15.00.

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5 hours = $5.00

10 hours = $10.00

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3 years ago

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.

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3 years ago