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AfilCa [17]
3 years ago
9

Jerry is trying to earn two hundred nine dollars for some new video games. if he charges forty-seven dollars to mow a lawn, how

many lawns will he need to mow to earn the money?
Mathematics
1 answer:
qwelly [4]3 years ago
4 0

We could represent this as an equation:

47x = $209, where x is the number of lawns mowed.

What we could do is divide 209 by 47 to find out how many lawns Jerry would need to mow. This would give us about 4.4 (when rounded). Since Jerry has to mow entire lawns, we would have to round that up to 5 lawns. Jerry would earn $235 for those, which is more than enough for those new video games.

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What is the area of a rectangle with side lengths of 1/2 and 2/3
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Answer:

2/6 square units

Step-by-step explanation:

What is the area of a rectangle with side lengths of 1/2 and 2/3

Step one:

given data

L= 1/2 units

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Step two:

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8 0
2 years ago
For Each Data Set, Determine the median, the first and third Quartiles, and the interquartile Range.
stiks02 [169]
1. First of all arrange the data set in either ascending or descending order.
12, 19, 24, 26, 31, 38, 53.  N = 7 (number of data items)
Median position = 1/2(N + 1)th item = 1/2(7 + 1)th item = 1/2(8)th item = 4th item = 26
First quatile = 1/4(N + 1)th item = 1/4(7 + 1)th item = 1/4(8)th item = 2nd item = 19
Third quatile = 3/4(N + 1)th item = 3/4(7 + 1)th item = 3/4(8)th item = 6th item = 38
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5 0
3 years ago
Can somebody please answer this question?
Trava [24]

Answer:

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Step-by-step explanation:

4 0
2 years ago
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The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing to d
NikAS [45]

Answer:

Option b. should not be rejected

Step-by-step explanation:

We are given that the contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces.

We have to test whether the variance of the population is significantly more than 0.003, i.e.;

  Null Hypothesis, H_0 : \sigma = \sqrt{0.003}

Alternate Hypothesis, H_1 : \sigma > \sqrt{0.003}

The test statistics used here for testing variance is;

          T.S. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2}__n_-_1

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           n = sample size = 26 cans

So, Test statistics = \frac{(26-1)0.06^{2} }{0.003 } ~ \chi^{2}__2_5

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So, at 5% level of significance chi square table gives critical value of 37.65 at 25 degree of freedom. Since our test statistics is less than the critical so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that null hypothesis should not be rejected and variance of population is 0.003.

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