Round 8.795 to the nearest cent

The mean of the commute time to work for a resident of a certain city is 27.3 minutes. Assume that the standard deviation of the

posledela

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Mean commute time (m) = 27.3 minutes

Standard deviation (sd) = 7.1 minutes

a) What minimum percentage of commuters in the city has a commute time within 2 standard deviations of the mean?

Using chebyshev's theorem ;

No more than 1/k² values of a distribution can be

k standard deviations from the mean.

Here k = 2

Hence,

1/k² = 1/2² = 1/4 = 0.25

Hence, minimum percentage of commmuters within 2 standard deviations of the mean :

1 - 0.25 = 0.75 m

B.) What minimum percentage of commuters in the city has a commute time within 1.5 standard deviations of the mean?

Here, k = 1.5

1/k² = 1/1.5² = 1/2.25 = 0.444

Hence, minimum commute time within 2 standard deviations of the mean :

1 - 0.444 = 0.56 = 56%

Commute time within 1.5 standard deviation

[27.3 - (1.5 × 7.1), 27.3 + (1.5 * 7.1)

[16.65, 37.95]

C.) What is the minimum percentage of commuters who have commute times between 6minutes and 48.6 minutes?

X - m /sd

X = 6

= (6 - 27.3) / 7.1 = - 3

X = 48.6

= (48.6 - 27.3) / 7. 1 = 3

Hence,

1/k² = 1 / 3² = 1/9

1 - 1/9 = 8/9

= 0.888 = 0.89 = 89%

3 0

4 years ago

True or False

Hitman42 [59]

Answer:

True

Step-by-step explanation:

To find the average of anything, you add their values together, and then divide that number by how many of them you added.

Here, you are trying to calculate monthly expenditure, so you add their values together (the sum of monthly expenditures) and then you divide it by how many months you used to add up to the sum (divided by the number of months).

Hope this helped ya! :)

4 0

3 years ago

Find so that the distance between (−2,3) and (,1) is √13

belka [17]

The distance between (-2, 3) and (-5, 1) is √13.

or, the distance between (-2, 3) and (1, 1) is √13.

We know that the length of the line segment connecting any two points represents the distance between them. There is just one line that connects the two points. Therefore, by measuring the length of the line segment that connects the two points, the distance between them can be determined. If (a, b) and (c, d) be two points, then the distance between them is $\sqrt[]{(b - a)^{2} +(d- c)^{2} }$ .

Here, one point is (-2, 3).

Let the other point be (x, 1).

Given that the distance is √13.

Now, $\sqrt[]{(x - (-2))^{2} +(1 - 3)^{2} } = \sqrt{13}$

i.e. $\sqrt[]{(x + 2)^{2} +( - 2)^{2} } =\sqrt{13}$

i.e. $\sqrt[]{x^{2}+4x +4 +4 }=\sqrt{13}$

i.e. $x^{2}+4x +8 =13$

i.e. $x^{2}+4x + 8- 13=0$

i.e. $x^{2}+4x -5=0$

i.e. $x^{2} +5x - x -5=0$

i.e. $x(x+5)-1(x+5)=0$

i.e. $(x+5)(x-1)=0$

i.e. $x=-5,1$

So, the point is either (-5, 1) or (1, 1).

Therefore, the required point is either (-5, 1) or (1, 1).

i.e. the distance between (-2, 3) and (-5, 1) is √13.

or, the distance between (-2, 3) and (1, 1) is √13.

Learn more about distance here -

brainly.com/question/431605

#SPJ10

7 0

2 years ago

A Jerome's rock band recorded 13 songs during an afternoon recording session. Each song took 1/6 of a side of a tape to record.

Lapatulllka [165]

Answer:
number of tape sides = 2.1667 tape sides

Explanation:
We know that each song took 1/6 of a side of tape to be recorded.
To know how many tape sides were used to record 13 songs, all we have to do is cross multiplication as follows:
1 song ..................> 1/6 of a tape side
13 songs ..............> ?? tape sides

Number of tape sides = [(13) * (1/6)] / 1
number of tape sides = 2.1667 tape sides

Hope this helps :)

8 0

4 years ago

Given: Which additional congruence statement could you use to prove that ∆BJK ≅ ∆CFH by the HL Theorem? ∠BJK ≅ ∠CFH ∠B ≅ ∠C

earnstyle [38]

Answer:

∠BJK ≅ ∠CFH

Step-by-step explanation:

∆BJK ≅ ∆CFH then we say ∠BJK ≅ ∠CFH

Only when we say ∠BJK ≅ ∠CFH we can say (expand) and speak of lengths by comparing two of each ∠BJA ≅ ∠AFC whilst ∠BK≅ ∠ HC then repeat with∠BKJ ≅ ∠CHF etc.

3 0

3 years ago

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