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Alex
3 years ago
12

Your crew can pave two driveways in a day. You work five days per week but take off 2 weeks each year for holiday and vacation.

How many driveways could your crew pave in a year?
Mathematics
2 answers:
Llana [10]3 years ago
7 0
500 hope this helped :))
soldier1979 [14.2K]3 years ago
4 0

Answer:

500


Step-by-step explanation:

Crew can pave 2 driveways in 1 day.

Thus, crew can pave 2*5=10 driveways in 1 week (5 days work week).

<u><em>In total there are 52 weeks in a year. But 2 weeks in a year are for vacation, so 50 weeks in total for a year.</em></u>

Thus in 50 weeks @ 10 driveways per week, the crew can pave:

50*10=500 driveways.

So, the crew can pave 500 driveways in a year.

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125 divided by 5 is 25. hope this helped but im not to sure what friendly parts are sorry
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The two pictures are the questions and question five is what i need help with.
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Answer:yes they do

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Elaine mellon earns $17.80 an hour as an electronic billing specialist for a legal clinic. she must work ____ hours a week, to e
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37.9 hours

Which is 38 hours after rounding it.

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3 years ago
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Solve for x<br>A:10 B:14.5 C:20 D:6÷3​
den301095 [7]

By applying Cathetus theorem, the value of x is equal to 10 units.

<h3>How to determine the value of x?</h3>

In order to determine the value of x, we would apply Cathetus theorem (leg rule), which states that each leg of a right-angled triangle is the geometric mean that's directly proportional between the hypotenuse and the part of the hypotenuse that's directly below the leg.

In this context, we have:

x² = m × a

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x² = 25 × 4

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Read more on Cathetus theorem here: brainly.com/question/11357448

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7 0
1 year ago
The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated
cluponka [151]

Answer:

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

<em>Let </em>p_1<em> = proportion of cockroaches that died on glass surface.</em>

<em />p_2<em> = proportion of cockroaches that died on plasterboard surface.</em>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here <u>Two-sample z proportion</u> <u>statistics</u>;

                       T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}  } }  ~ N(0,1)

where, \hat p_1 = sample proportion of cockroaches that died on glass surface = \frac{18}{36} = 0.50

\hat p_2 = sample proportion of cockroaches that died on plasterboard surface = \frac{25}{36} = 0.694

n_1 = sample of cockroaches on glass surface = 36

n_2 = sample of cockroaches on plasterboard surface = 36

So, <u><em>test statistics</em></u>  =  \frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36}  } }

                               =  -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

8 0
3 years ago
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