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Vikki [24]
4 years ago
9

Three students are arguing over who can run the fastest. Student A says that she can run 2/3 mile in six minutes. Student B says

that he can run 1/2 mile in four minutes. Student C says that he can run 4/5 mile in eight minutes. What fraction of a mile can Student C run in one minute?
Mathematics
1 answer:
nordsb [41]4 years ago
3 0
1/10 of a mile. 4/5 is equivalent to .8, and dividing that by 8 would give you .1, and that is equivalent to 1/10.
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Korvikt [17]
The answer for 378 x 9 = 3402
7 0
4 years ago
(Based on Q1 ~ Q3) According to the Bureau of the Census, 18.1% of the U.S. population lives in the Northeast, 21.9% inn the Mid
vekshin1

Answer:

We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.

Step-by-step explanation:

The null Hypothesis: Geographical distribution of hotline callers could be the same as the U.S. population distribution

Alternative hypothesis: Geographical distribution of hotline callers could not be the same as the U.S. population distribution

The populations considered are the Midwest, South, Northeast, and west.

The number of categories, k = 4

Number of recent calls = 200

Let the number of estimated parameters that must be estimated, m = 0

The degree of freedom is given by the formula:

df = k - 1-m

df = 4 -1 - 0 = 3

Let the significance level be, α = 5% = 0.05

For  α = 0.05, and df = 3,

from the chi square distribution table, the critical value = 7.815

<u>Observed and expected frequencies of calls for each of the region:</u>

<u>Northeast</u>

Observed frequency = 39

It contains 18.1% of the US Population

The probability = 0.181

Expected frequency of call = 0.181 * 200 = 36.2

<u>Midwest</u>

Observed frequency = 55

It contains 21.9% of the US Population

The probability = 0.219

Expected frequency of call = 0.219 * 200 =43.8

<u>South</u>

Observed frequency = 60

It contains 36.7% of the US Population

The probability = 0.367

Expected frequency of call = 0.367 * 200 = 73.4

<u>West</u>

Observed frequency = 46

It contains 23.3% of the US Population

The probability = 0.233

Expected frequency of call = 0.233 * 200 = 46

x^{2} = \sum \frac{(O_{i} - E_{i})  ^{2} }{E_{i} } ,   i = 1, 2,.........k

Where O_{i} = observed frequency

E_{i} = Expected frequency

Calculate the test statistic value, x²

x^{2} = \frac{(39 - 36.2)^{2} }{36.2} + \frac{(55 - 43.8)^{2} }{43.8} + \frac{(60 - 73.4)^{2} }{73.4} + \frac{(46 - 46.6)^{2} }{46.6}

x^{2} = 5.535

Since the test statistic value, x²= 5.535 is less than the critical value = 7.815, the null hypothesis will not be rejected, i.e. it will be accepted. We can therefore conclude that the geographical distribution of hotline callers could be the same as the U.S population distribution.  

7 0
3 years ago
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The answer is 7 and 4/ 7+4+11 and 7-4=3!
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For this question it would look like this 4/3 * 3.14 * 8^3=v

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Which explanation uses the number line to show 5 + (-5) =0
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Answer:

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Step-by-step explanation:

I. If 5 + (-5) = 0

  then 5 - 5 = 0

  If 5 is on the right side

   then -5 is on the left side

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3 years ago
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