State whether the following are describing distance (d), velocity (v), or acceleration (a)

A Sum of $14 is made of 92 coins consisting of dimes and quarters. How Many are there of each kind of coin

steposvetlana [31]

So first I would say, what if all of them were dimes, how far away would it be from $14?

So 92 coins * 10 cents = $9.20

So it's 4.80 dollars away from 14 dollars.

So if we were to switch one to a quarter, it would increase by 0.15 cents.

So we want to see how many increases we need to reach 4.80 dollars more.

4.80/0.15 = 32

So there are 32 quarters and 60 dimes.

7 0

3 years ago

Read 2 more answers

The perimeter of a rectangle is 22cm. the area is 28cm2. what are the lengths of the sides ?

MissTica

Answer: length: 7cm

width:4cm

Step-by-step explanation:

Area: L × W

7×4=28

Perimeter =L±L+W+W

7+7+4+4=22

7 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:

$\bar x = 70.2903$