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Vitek1552 [10]
3 years ago
15

A gazelle runs 780 feet in 12 seconds. Angela wants to determine how far a gazelle could run in 1 minute at this rate. Explain h

ow to use equivalent fractions to find the distance a gazelle could run in 1 minute. Then write the equivalent fractions.
Mathematics
1 answer:
Hitman42 [59]3 years ago
4 0

Answer:

In 1 min a gazelle runs 3,900 ft.

As a fraction, 1/65

In order to find the solution to this problem you could use a table.

Since we are using seconds we have to convert the 1 min = 60 seconds.

<u>Sec</u><u>   |  12    |    60</u>

<u>Fts</u><u>    |  780  |    x  </u>

12 goes into 60 5 times, so we have to multiply 780 by 5 which gives us 3900.

x=3900

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Newborn babies: A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights
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Answer:

a) 615

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Step-by-step explanation:

According to the Question,

  • Given that,  A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams

  • Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.

Z = (x - mean)/standard deviation

Now,

For x = 4171,  Z = (4171 - 3311)/860 = 1  

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Next, multiply that by the sample size of 732.

  • Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171  

 

  • For part b, use the same method except x is now 1591.    

Z = (1581 - 3311)/860 = -2    

  • P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.

 

  • For part c, we now need to get two Z scores, one for 3311 and another for 5031.

Z1 = (3311 - 3311)/860 = 0

Z2 = (5031 - 3311)/860= 2  

P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772

  approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.

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Step-by-step explanation:

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