Математика срочно пожалуйста

kirill115 [55]

Answer:

I believe it is the top right scatterplot, I can't tell which letter, I think B though.

Step-by-step explanation:

If you were to put a line of best fit for each graph, the top right scatterplot would be the only one that has a straight line of best fit.

Hope this helps :)

3 0

2 years ago

Read 2 more answers

Mrs. Johnson gets her morning coffee. She can't decide which size to order. Which size is the better buy? Explain.

Tom [10]

Answer:

large

Step-by-step explanation:

3.95 divided by 12 = 0.33

4.25 divided by 16 = 0.27

4.55 divided by 20 = 0.2275

the 20oz is the least unit price

3 0

2 years ago

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-

Tresset [83]

Answer:

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

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:

$\bar x = 70.2903$