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Basile [38]
3 years ago
11

umber line to answer Problems '1' ed 360 cans and bones while fundraising fur hols baseball team This was of what Raul Haw many

cans and bottles did Reggie collect

Mathematics
1 answer:
Anastasy [175]3 years ago
5 0
The question is if 360 is 40%, what is the 100% (Reggies), so we solve it using a rule of three:
 40% -----> 360
100% -----> x
x = (100)(360)/40
x = 900
Therefore the 100% is 900, so Reggie collected 900 cans
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a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-
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\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

M = \frac{1}{2}(Lower + Upper)

Where

Lower \to Lower class interval

Upper \to Upper class interval

So, we have:

Class 63-65:

M = \frac{1}{2}(63 + 65) = 64

Class 66 - 68:

M = \frac{1}{2}(66 + 68) = 67

When the computation is completed, the frequency distribution will be:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

\bar x = 70.2903

\sigma = 3.5795

<em>See attachment for result of 1-VarStats</em>

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