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Anuta_ua [19.1K]
2 years ago
7

Amy runs 4 miles in 31 minutes how many minutes does she take per mile

Mathematics
1 answer:
Hitman42 [59]2 years ago
4 0

Answer:

about 8 minutes per mile

Step-by-step explanation:

a= t/d

31/4=  7.75

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Don’t know what the answer is
Lyrx [107]

Answer:

210

Step-by-step explanation:

if you divide 126 by 3 you get 42 than multiply by 5 and you get 210

5 0
3 years ago
The rectangle below has an area of zº +82 + 15 square meters and a width of 2 + 3 meters.
Lelu [443]

Answer:

(x + 5) m

Step-by-step explanation:

From the question given above, the following data were obtained:

Area (A) = (x² + 8x + 15) m²

Width (W) = (x + 3) m

Length (L) =?

The lenght of the rectangle can be obtained as follow:

A = L × W

x² + 8x + 15 = L (x + 3)

Divide both side by x + 3

L = x² + 8x + 15 / x + 3

L = (x + 5) m

Please attached photo for explanation.

Therefore, the length of the rectangle is: (x + 5) m

8 0
3 years ago
2587 to the nearset thousand
BigorU [14]

Answer:

3000

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
7 pounds of rice is 4$ how much is one pound of rice?​
blagie [28]

Answer:

$0.57

Step-by-step explanation:

7 pound = 4 $

1 pound =4/7 $

=0.57$

4 0
1 year ago
Read 2 more answers
A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

                    Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

                                                 = 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0
3 years ago
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