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1 year ago

Mrs Davis worked 42 hr 35 min last week how many hours and minutes did she average in each of the five days

Mathematics

2 answers:

DerKrebs [107]1 year ago

6 0

Answer:

8 hours and 31 minutes

Step-by-step explanation:

To get an average, you take the total time and divide it by the total days. Here, it would be 42 hours and 35 minutes divided by 5 days. You can convert the hours to minutes, and get 42x60=2520 plus the 35 minutes to get 2555 minutes. Divise that by 5 and you get 511 minutes per day. Since they're asking for hours and minutes, divide 511 by 60 to get 511÷60=8 with remainder 31.

gayaneshka [121]1 year ago

3 0

Answer:

8 hours and 31 minutes per day

Step-by-step explanation:

$\implies 42 \text{hours} \times 60\text{minutes}=2,250 \text{minutes}$

$\implies2,520 +35 = 2,555$

$\implies 2,555 /5 = 511$

$\implies 511/60 = 8 \frac{31}{60} = 8\text{hours}, 31 \text{minutes per day}$

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Andrews [41]

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3 years ago

Read 2 more answers

Probability of father and mother born in Michigan independent.

raketka [301]

Answer:

A. no, not independent

Step-by-step explanation:

Let A = "father born in Michigan" and B = "mother born in Michigan".

If A and B are independent, then ...

P(A|B) = P(A)

The table tells us ...

P(A|B) = P(A&B)/P(B) = 0.29/0.73 ≈ 0.397

and

P(A) = 0.48

Since 0.48 ≠ 0.397, we must conclude that the events are not independent.

6 0

2 years ago

Simplify the expression 8x + 4(x -1)

s344n2d4d5 [400]

8x+4x-4

12x-4 is the answer

6 0

3 years ago

Carson purchased supplies for his computer for $64. The sales tax was $4.16. He gave the salesperson $80. How much change did he

krek1111 [17]

To work this out you need to do 80 - 64 - 4.16 = 11.48

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2 years ago

The U.S. Census Bureau conducts annual surveys to obtain information on the percentage of the voting-age population that is regi

yan [13]

Answer:

We conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

Step-by-step explanation:

We are given that 513 employed persons and 604 unemployed persons are independently and randomly selected, and that 287 of the employed persons and 280 of the unemployed persons have registered to vote.

Let $p_1$ = percentage of employed workers who have registered to vote.

$p_2$ = percentage of unemployed workers who have registered to vote.

So, Null Hypothesis, $H_0$ : $p_1\leq p_2$ {means that the percentage of employed workers who have registered to vote does not exceeds the percentage of unemployed workers who have registered to vote}

Alternate Hypothesis, $H_A$ : $p_1>p_2$ {means that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote}

The test statistics that would be used here Two-sample z test for proportions;

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 employed workers who have registered to vote = $\frac{287}{513}$ = 0.56

$\hat p_2$ = sample proportion of unemployed workers who have registered to vote = $\frac{280}{604}$ = 0.46

$n_1$ = sample of employed persons = 513

$n_2$ = sample of unemployed persons = 604

So, the test statistics = $\frac{(0.56-0.46)-(0)}{\sqrt{\frac{0.56(1-0.56)}{513}+\frac{0.46(1-0.46)}{604} } }$

= 3.349

The value of z test statistics is 3.349.

Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.

Since our test statistic is more than the critical value of z as 3.349 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

5 0

2 years ago