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cestrela7 [59]
2 years ago
6

A 16\dfrac1216 2 1 ​ 16, start fraction, 1, divided by, 2, end fraction kilometer stretch of road needs repairs. Workers can rep

air 2\dfrac142 4 1 ​ 2, start fraction, 1, divided by, 4, end fraction kilometer of road per week. How many weeks will it take to repair this stretch of road?
Mathematics
1 answer:
Solnce55 [7]2 years ago
7 0

Answer:

It will take  7\frac{1}{3} weeks to complete the repair of the stretch of road.

Step-by-step explanation:

Given:

Length of the roads that need repairs = 16\frac{1}{2}\ km

Length of the road workers can repair in a week = 2\frac{1}{4}\ km

To find the number of weeks will it take to repair this stretch of road.

Solution:

Unit rate of workers to repair  roads =   2\frac{1}{4}\ km per week.

Total length of the roads to repair = 16\frac{1}{2}\ km

The number of weeks it will take to repair  16\frac{1}{2}\ km of road by workers can be given as:

⇒ \frac{16\frac{1}{2}\ km}{2\frac{1}{4}\ km}

<em>In order to divide mixed numbers, we convert them first to fractions.</em>

⇒ \frac{\frac{33}{2}} {\frac{9}{4}}

<em>To divide fractions, we take reciprocal of the divisor and replace division with multiplication:</em>

⇒  \frac{33}{2}\times \frac{4}{9}

⇒ \frac{22}{3}

We reconvert the fraction to mixed number.

⇒ 7\frac{1}{3} weeks

Thus, it will take  7\frac{1}{3} weeks to complete the repair of the stretch of road.

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Answer:

The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).

Step-by-step explanation:

Intern            No. of Breast

Number        Exams Performed               X²

1                         30                                  900

2                        40                                 1600

3                        8                                        64

4                        20                                   400

5                          26                                 676

6                           35                               1225

7                            35                               1225

8                            20                                400

9                             25                              625

<u>10                                 20                        400 </u>

<u>                                   </u><u> ∑ 259                  ∑ 7515</u>

Mean= X`= ∑x/n= 259/10= 25.9

Variance = s²= 1/n-1[∑X²- (∑x)²/n]

= 1/0[7515- (259)²/10]= 1/9[7515- 6708.1]

= 806.9/9=89.655= 89.66

Standard Deviation= √89.655= 9.4687

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The value of t with significance level alpha= 0.05 and 9 degrees of freedom  is t(0.025,9)= 2.262

The 95 % Confidence interval is given by

x`±t(∝,n-1) s/√n

So Putting the values

25.9± 2.262( 9.4687/√10)

= 25.9 ±2.262 (2.9943)

= 25.9 ± 6.7730

= 25.9 +6.7730=32.6730

25.9 -6.7730= 19.1269

= 19.1269, 32.6730

The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).

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