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goldfiish [28.3K]
3 years ago
12

Mark runs back and forth on a football field for a total of 3000 yards. How many miles did Mark run to the nearest tenth of a mi

le?
Mathematics
2 answers:
galben [10]3 years ago
7 0

Answer:

the answer is 1.7 miles

Step-by-step explanation:

gizmo_the_mogwai [7]3 years ago
5 0

Answer:

1.7 miles.

Step-by-step explanation:      

We have been given that Mark runs back and forth on a football field for a total of 3000 yards.

Since we know that 1 mile equals to 1760 yards. To find the number of miles Mark ran we will divide 3000 yards by number of yards in one mile (1760).

\text{ Mark ran}=\frac{3000\text{ yards}}{1760\frac{\text{yards}}{\text{mile}}}

\text{ Mark ran}=\frac{3000\text{ yards}}{1760}\times \frac{\text{miles}}{\ext{yards}}

After cancelling out yards from numerator and denominator we will get,

\text{ Mark ran}=\frac{3000}{1760} \text { miles}}

\text{ Mark ran}=1.7045454545454545\text{ miles}}\approx 1.7\text { miles}}

Therefore, Mark ran 1.7 miles on the football field.          

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Using a proportion, what can you infer about the number of households in her town that have more than three children?
eduard

Answer:

About 296 households have more than three children.

Step-by-step explanation:

The complete question is:

Carmen used a random number generator to simulate a survey of how many children live in the households in her town. There are 1,346 unique addresses in her town with numbers ranging from zero to five children. The results of 50 randomly generated households are shown below. Children in 50 Households Number of Children Number of Households 0 5 1 11 2 13 3 10 4 9 5 2 Using a proportion, what can you infer about the number of households in her town that have more than three children? About 242 households have more than three children. About 269 households have more than three children. About 296 households have more than three children. About 565 households have more than three children.

Solution:

The data provided for the number of children and number of households is:

Number of Children (<em>X</em>)          Number of Households (<em>f</em> (X))

               0                                                   5

               1                                                    11

               2                                                   13

               3                                                   10

               4                                                    9

               5                                                    2

           TOTAL                                             50

Compute the probability of households having more than three children as follows:

P (More than 3 children) = P (4 children) + P (5 children)

                                        =\frac{9}{50}+\frac{2}{50}

                                        =\frac{11}{50}

                                        =0.22

Let the random variable <em>Y</em> be defined as the number of households having more than three children.

The probability of the random variable <em>Y </em>is, <em>p</em> = 0.22.

The entire population, i.e. total number of addresses in Carmen's town with numbers ranging from zero to five children, consists of <em>N</em> = 1,346 households.

Compute the expected number of households having more than three children as follows:

E(Y) =N\times p

         =1346\times 0.22\\=296.12\\\approx 296

Thus, about 296 households have more than three children.

4 0
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