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xeze [42]
3 years ago
7

A student athlete run 3 1/3 miles in 30 minutes A professional runner can run 1 1/4 times as far in 30 minutes . how far can the

professional runner run in 30 minutes
Mathematics
1 answer:
IrinaK [193]3 years ago
4 0
The professional runner can run 4 1/6 miles in 30 minutes.

Since the professional can run 1 1/4 times as far in 30 minutes, we multiply 3 1/3 by 1 1/4:

(3 1/3)(1 1/4)

Convert each to an improper fraction (multiply the whole number by the denominator and add the numerator):
10/3(5/4) = 50/12 = 25/6 = 4 1/6.
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Finding the volume of the solid.Use unit cubes to help.
9966 [12]

Answer:

the answer is 125. if you cut them slice to slice you'll get 25 because of there five slices of 25 all you got to do is 25 x 5 and you'll get 125.

Hope this help

Have a nice day :)

4 0
3 years ago
What is the value of y in the solution to the system of equations?
Molodets [167]

Answer: y=32/5

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Consider this expression.
schepotkina [342]

Answer:

9m^2 + 17m- 9

Step-by-step explanation:

Given

7m^2 + (2m - 1)(m + 9)

Required:

Simplify

We start by opening the bracket

7m^2 + (2m(m + 9) - 1(m + 9))

7m^2 + 2m^2 + 18m - m - 9

Collect Like Terms

9m^2 + 17m- 9

Hence, the equivalent expression is 9m^2 + 17m- 9

5 0
3 years ago
A number line ranging from 5 to 45 with a box encompassing the numbers 20, 25, and 30.
irina1246 [14]

Answer:

the maximum value is 45

Step-by-step explanation:

  • A box of leaf plot has 5 points and they are:

The minimum, the first quartile (Q1) the median (Q2), the third quartile (Q3), and the maximum. The box encompasses Q1, Q2, and Q3. The maximum is after Q3 and the minimum is before Q1.

In this case,

Q1, Q2, and Q3 are in a range 20-30, thus the maximum come after 30, which is 45

4 0
2 years ago
Suppose the heights of 18-year-old men are approximately normally distributed, with mean 65 inches and standard deviation 4 inch
Serhud [2]

Answer:

Step-by-step explanation:

<u>a)</u>

  • Given that ; X ~ N ( µ = 65 , σ = 4 )
  • P ( 64 < X < 66 )

From application of normal distribution ;

  • Z = ( X - µ ) / σ, Z = ( 64 - 65 ) / 4,  Z = -0.25
  • Z = ( 66 - 65 ) / 4,  Z = 0.25

Hence, P ( -0.25 < Z < 0.25 ) = P ( 64 < X < 66 ) = P ( Z < 0.25 ) - P ( Z < -0.25 ) P ( 64 < X < 66 ) = 0.5987 - 0.4013

  • P ( 64 < X < 66 ) = 0.1974

b) X ~ N ( µ = 65 , σ = 4 )

  • P ( 64 < X < 66 )

From normal distribution application ;

  • Z = ( X - µ ) / ( σ / √(n)), plugging in the values,
  • Z = ( 64 - 65 ) / ( 4 / √(12)) = Z = -0.866
  • Z = ( 66 - 65 ) / ( 4 / √(12)) = Z = 0.866

P ( -0.87 < Z < 0.87 )

  • P ( 64 < X < 66 ) = P ( Z < 0.87 ) - P ( Z < -0.87 )
  • P ( 64 < X < 66 ) = 0.8068 - 0.1932
  • P ( 64 < X < 66 ) = 0.6135

c) From the values gotten for (a) and (b), it is indicative that the probability in part (b) is much higher because the standard deviation is smaller for the x distribution.

4 0
3 years ago
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