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maksim [4K]
3 years ago
6

On the first play, the Wobegone High School football team lost 2 yards. On each of the next 3 plays they lost 2 yards. How many

yards did they total on the 4 plays?
Mathematics
1 answer:
inessss [21]3 years ago
8 0
So if they lost 2 yards on the first play and on the following three plays they lost 2 yards they  lost 4 yards total because it doesnt say they lost 2 yards on each of the three play so they would only lose 4 yards 
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I got 11.5 

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What do we call moving a record back and forth in a rhythm?
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25 points:)<br><br> question 2-3
Furkat [3]

Answer:

2) r=73

Step-by-step explanation:

all 4 sided shapes add up to 360.

so 90+90=180

180+107=287

360-287=73

so r=73

4 0
3 years ago
What is the present value of 6000 received 14 years from today when the interest rate is 10%
likoan [24]

Answer:

$14400

Step-by-step explanation:

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Principal= 6000

Time= 14 years

Rate= 10%

Let us use the simple interest expression to find the final amount

A=P(1+rt)

substitute

A=6000(1+0.1*14)

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3 0
2 years ago
In a test of the effectiveness of garlic for lowering​ cholesterol, 8181 subjects were treated with raw garlic. Cholesterol leve
xz_007 [3.2K]

Answer:

With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

Step-by-step explanation:

The dependent <em>t</em>-test (also known as the paired <em>t</em>-test or paired samples <em>t</em>-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case a paired <em>t</em>-test is used to determine the effectiveness of garlic for lowering​ cholesterol.

A random sample of 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment.

The hypothesis for the test can be defined as follows:

<em>H₀</em>: With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0, i.e. <em>d</em> ≤ 0.

<em>Hₐ</em>: With garlic​ treatment, the mean change in LDL cholesterol is greater than 0, i.e. <em>d</em> > 0.

The information provided is:

\bar d=0.40\\SD_{d}=16.2\\\alpha =0.01

Compute the test statistic value as follows:

t=\frac{\bar d}{SD_{d}/\sqrt{n}}\\\\=\frac{0.40}{16.2/\sqrt{81}}\\\\=0.22

The test statistic value is 0.22.

Decision rule:

If the <em>p</em>-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the <em>p</em>-value of the test as follows:

p-value=P(t_{n-1}>0.22)\\=P(t_{80}>0.22)\\=0.4132

*Use a <em>t</em>-table.

The <em>p</em>-value of the test is 0.4132.

<em>p-</em>value= 0.4132 > <em>α</em> = 0.01

The null hypothesis was failed to be rejected.

Thus, it can be concluded that with garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

3 0
2 years ago
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