If 140 bricks are used to build a wall 20 feet high, how many bricks will you use to build a wall 30 feet high?

Kim rode her bicycle 135 miles in 9 weeks riding the same distance each week. eric rode his bicycle 102 miles in 6 weeks riding

zysi [14]

The answer is a because you divide 135÷9=15 and 102÷6=17 . 17-15=2

6 0

3 years ago

If y = 8x − 2, which of the following sets represents possible inputs and outputs of the function, represented as ordered pairs?

Nataly [62]

You just need to try with all of the pair f.ex
0,-2 -> put 0 in the x-> y=8*0-2=-2 so ok it is q possible pair of the function..
You try every pair till you find set where all the pairs fit
So set 1
1 and 6: 8*1-2=6 ok
2 and 14 : 8*2-2=14 ok
Hope I could help

6 0

3 years ago

Read 2 more answers

A major convention center, whose floors are covered by a square carpet tiles with side lengths of 9 inches laid out in a 48 by 9

bearhunter [10]

Answer:

option B

$28.80

Step-by-step explanation:

Given the question,

side length of one square tile = 9 inches

area of square tile = l² = 9² = 81 inches²

There are 48(96) = 4608 carpet tiles, so the company was making 4608($.05) = $230.40.

Now, the company countered with square feet. We have a total of 4608(81) = 373,248 square inches.

1 square foot = 144 square inches.

373,248/144 = 2592 square ft.

2592($.10) = $259.20

$259.20 - $230.40 = $28.80

4 0

3 years ago

3+3+7+8+5+4+33yx11+<br><br><br> COME on People easy math for 30points.

Paraphin [41]

Answer:

363y + 30

Step-by-step explanation:

3 0

3 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:

$\bar x = 70.2903$

$\sigma = 3.5795$

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

8 0

3 years ago