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kkurt [141]
3 years ago
10

Furry can eat 10 mangoes in 5 minutes. How long does it take Furry to eat 18 mangoes at the same speed?

Mathematics
2 answers:
Ad libitum [116K]3 years ago
6 0
9 minutes

30 seconds each mango

White raven [17]3 years ago
6 0
It would be 8 minutes


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working together, melissa and jing can mow a lawn in 5 hours. It would take melissa 8 hours to do the job alone. What is the val
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Total Time = (M x K) / (M + K)
5 = (8 x K) / (8 + K)
40 + 5 K = 8K
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King can do the job in 13.33 Hours

King could mow (1 / 13.33) of the lawn in one hour.

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3 years ago
Which fraction is equivalent to 15/30<br> A 3/4<br> B 2/4<br> C 3/5<br> D 5/6
Ratling [72]

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2/4

Step-by-step explanation:

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100 points hurry<br> given the angle is 38 what is the angle measure of its vertical angle
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Which is the best estimate of x based on rounding the constants and coefficients in the equation to the nearest integer? 6.2x+1.
DerKrebs [107]

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</span>estimate<span>
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7 0
3 years ago
Listed below are the measured radiation absorption rates? (in w/kg) corresponding to 11 cell phones. Use the given data to const
deff fn [24]

Answer:

The minimum value is 0.51

Q_1= \frac{0.89+1.04}{2}= 0.965

Median=Q2= 1.18

Q_3= \frac{1.41+1.42}{2}= 1.415

Max = 1.49

Step-by-step explanation:

We assume the following dataset:

1.18,1.41, 1.49,1.04,1.04,0.74,0.89,1.42,1.45,0.51,1.38

We order the dataset on increasing way and we have:

0.51 0.74 0.89 1.04 1.04 1.18 1.38 1.41 1.42 1.45 1.49

The minimum value is 0.51

The first quartile Q1 can be calculated with the first 6 observations: 0.51 0.74 0.89 1.04 1.04 1.18. The Q1 would be the average between the 3rd and 4th position:

Q_1= \frac{0.89+1.04}{2}= 0.965

For the median since the number of data represent an odd number than the median would be the position in the middle (6th)

Median=Q2= 1.18

The first quartile Q1 can be calculated with the last 6 observations: 1.18 1.38 1.41 1.42 1.45 1.49. The Q1 would be the average between the 3rd and 4th position:

Q_3= \frac{1.41+1.42}{2}= 1.415

And the maximum value for this case would be:

Max = 1.49

We can see the boxplot obtained on the figure attached.

5 0
3 years ago
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