Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 47.3 miles per hour. Speed (miles per hour) 42-45 46-49 50-53 54-57 58-61Frequency 21 15 6 4 2The mean of the frequency distribution miles per hour. (Round to the nearest tenth as needed.)Which of the following best describes the relationship between the computed mean and the actual mean?A. The computed mean is not close to the actual mean because the difference between the means is more than 5%.B. The computed mean is not close to the actual mean because the difference between the means is more than 5%.C. The computed mean is not close to the actual mean because the difference between the means is less than 5%.D. The computed mean is close to the actual mean because the difference between the means is less than 5%.

47.4167

D. The computed mean is close to the actual mean because the difference between the means is less than 5%

Step-by-step explanation:

Given the distribution :

Speed(m/hr) ___ midpoint(x) ___ F ___fx

42 - 45 43.5 21 __ 913.5

46 - 49 47.5 15 __712.5

50 - 53 51.5 6 __ 309

54 - 57 55.5 4 ___ 222

58 - 61 59.5 2 ___ 119

The midpoint, x= sum of lower and upper boundary divided by 2

For instance, (42 + 45) / 2 = 43.5

The computed mean, Σfx / Σf = 2276 / 48 = 47.4167

Actual mean = 47.3 miles

(47.4167 - 47.3) / 47.3 * 100% = 0.24%

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: