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3 years ago

Kevin runs 8 miles in 84 minutes. At the same rate, how many minutes would he take to run 6 miles?

Mathematics

1 answer:

netineya [11]3 years ago

3 0

Kevin runs:

8 miles -- 84 minutes;

6 miles -- x minutes.

Write a proportion:

$\dfrac{8}{6}=\dfrac{84}{x}.$

Then $8\cdox =6\cdot 84,\\ \\x=\dfrac{6\cdot 84}{8}=3\cdot 21=63\ minutes.$

Answer: it takes him 63 minutes to run 6 miles

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LA plane is trying to travel 250 miles at a bearing of 20° E of S, however, it ends 230 miles away from the

Semmy [17]

Answer:

The wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ East of North with respect to the destination point.

Step-by-step explanation:

Let origin, O, br the starting point and point D be the destination at 250 miles at a bearing of 20° E of S, but due to wind let D' be the actual position of the plane at 230 miles away from the starting point in the direction of 35° E of South as shown in the figure.

So, we have |OD|=250 miles and |OD'|=230 miles.

Vector $\overrightarrow{DD'}$ is the displacement vector of the plane pushed by the wind.

From figure, the magnitude of the required displacement vector is

$|DD'|=\sqrt{|AB|^2+|PQ|^2}\;\cdots(i)$

and the direction is $\alpha$ east of north as shown in the figure,

$\tan \alpha=\frac{|PQ|}{|AB|}\;\cdots(ii)$

From the figure,

$|AB|=|OA-OB|$

$\Rightarrow |AB|=|OD\cos 20 ^{\circ}-OD'\cos 35 ^{\circ}|$

$\Rightarrow |AB|=|250\cos 20 ^{\circ}-230\cos 35 ^{\circ}|$

$\Rightarrow |AB|=45.52$ miles

Again, $|PQ|=|OP-OQ|$

$\Rightarrow |PQ|=|OD\sin 20 ^{\circ}-OD'\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=|250\sin 20 ^{\circ}-230\sin 35 ^{\circ}|$

$\Rightarrow |PQ|=46.42$ miles

Now, from equations (i) and (ii), we have

$|DD'|=\sqrt{|45.52|^2+|46.42|^2}=65.01$ miles, and

$\tan \alpha=\frac{|46.42|}{|45.52|}$

$\alpha=\tan^{-1}\left(\frac{|46.42|}{|45.52|}\right)=45.56 ^{\circ}$

Hence, the wind pushed the plane $65.01$ miles in the direction of $45.56 ^{\circ}$ E astof North with respect to the destination point.

5 0

3 years ago

Eliminate all exponents by Expanding 6^3 y^4

horrorfan [7]

Answer:

216*y*y*y*y

Step-by-step explanation:

6 cubed is 216, and y^4 expanded is yyyy. So if I'm understanding correctly, you want as your answer:

216*y*y*y*y

8 0

3 years ago

What does the relationship between the mean and median reveal about the shape of the data? The mean is less than the median, so

IgorC [24]

Answer:

The mean is equal to the median, so the data is symmetrical

Step-by-step explanation:

Here is the data.

10 5 8 10 12 6

8 10 15 6 12 18

The given data: 10 5 8 10 12 6 8 10 15 6 12 18

For finding the Mean, we will have to add all numbers together and divide it by total number. i.e sum of terms divided by number of terms

Mean= 10+5+8+10+12+6+8+10+15+6+12+18 ÷ 12

Mean = 120 ÷ 12 = 10

For finding the Median, first we need to rearrange the data in ascending order

5 6 6 8 8 10 10 10 12 12 15 18

We can see that the middle values are 10 and 10. So, the median will be the average of those two middle values.

Median = 10+10 ÷ 2

Median = 20 ÷ 2 = 10

From the calculation, we can see that both the median and mean are equal so, the data is symmetrical

5 0

3 years ago

Read 2 more answers

Brainliestis there is steps What is the solution to the equation log⁡〖 (2x+4)〗=2 ? Round to the nearest hundredth, if necessary.

Minchanka [31]

Answer:

x = 48

Step-by-step explanation:

log⁡〖 (2x+4)〗=2

Raise each side to the base of 10

10 ^log⁡〖 (2x+4)〗=10^2

The 10 ^log cancels

2x+4 = 100

Subtract 4 from each side

2x+4-4 = 100-4

2x = 96

Divide by 2

2x/2 = 96/2

x =48

3 0

3 years ago

The mean of the weights of a group of 100 men and women

Arte-miy333 [17]

Answer:

Step-by-step explanation:

160lb.

The mean of the weights of a group of 100 men and women is 160lb. If the number of men in the group is 60 and the mean weight of the men is 180lb, what is the mean weight of the women? For a set of data, the lower quartile is 19, the median is 31, and the upper quartile is 48.

7 0

2 years ago