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3 years ago

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3. All boys at a certain High school play basketball, tennis or volleyball. If 31% play basketball, and 45% play tennis, find th

e percentage of boys who play volleyball.

1 answer:

LekaFEV [45]3 years ago

7 0

24% is the answer I believe

You might be interested in

Janet can plant nine feet of carrots in 15 minutes while her daughter Amy can plant 17 feet of carrots in half an hour.

Lena [83]

Answer:

Step-by-step explanation:

In order to find who is more efficient at planting carrots, we want to put both ratios in terms of the same amount of time.

For simplicity, lets see how many carrots each can make in 1 hour.

$Janet:\\9ft \: in \: 15 \:minutes$

$\frac{9ft}{15min} = \frac{xft}{60min}\\x = \frac{60*9}{15} = 36$

Janet can plant 36 feet of carrots in one hour

$Amy:\\\frac{17ft}{30min} = \frac{xft}{60min}\\x = \frac{17*60}{30} = 34$

Amy can plant 34 feet of carrots in one hour

So, Janet can plant carrots more quickly

6 0

3 years ago

The ratio of a model building's dimensions to the building's actual dimensions is 1/52. If the height of the model is 15 inches,

g100num [7]

Set this up as a proportion with the model stuff on top and actual building stuff on bottom. $\frac{1}{52} = \frac{15}{x}$ . Cross multiply to get x = 780. The actual building is 780 inches. In feet that is 65.

4 0

3 years ago

Y^3-9y^2+y-9 I don't know how to solve this can you help

aleksley [76]

I'm guessing your problem is this:
y³ - 9y² + y - 9 = 0
right?

In solving this problem, I recommend doing this:
y³ - 9y² + y - 9 = 0
Factor out a y² from the first two numbers in the problem:
y²(y - 9) + (y - 9) = 0
Separate the parentheses which means y - 9 goes on one side. The y² added a one since it came from the + 1 in the middle of expression. When you're separating parentheses like this you just take the outside numbers and combine them together. Since + 1 came from the outside of the (y - 9) and y² also was sitting on the outside of (y - 9) combine them to make y² + 1. Like this:
(y² + 1)(y - 9) = 0
Now separate your two parentheses to two separate problems:
(y² + 1) = 0 and (y - 9) = 0
Now you're y² + 1 will equal:
y² = -1
y = √-1 <-- This number doesn't exist so it will be an imaginary number (i). If you guys didn't learn that in your class I recommend just leaving it as i for that part.
Now solve y - 9 = 0:
y = 9 <-- Since we added nine to both sides to get this.

So you're final answer should be y = i and 9

3 0

3 years ago

It takes 32 hours for a motorboat moving downriver to get from pier A to pier B. The return journey takes 48 hours. How long doe

lina2011 [118]

Answer:

Time taken by the un powered raft to cover this distance is T = 192.12 hr

Step-by-step explanation:

Let speed of boat = u $\frac{km}{hr}$

Speed of current = v $\frac{km}{hr}$

Let distance between A & B = 100 km

Time taken in downstream = 32 hours

$32 = \frac{100}{u +v}$

$u + v = \frac{100}{32}$

u + v = 3.125 ------ (1)

Time taken in upstream = 48 hours

$48 = \frac{100}{u -v}$

$u -v = \frac{100}{48}$

u - v = 2.084 ------- (2)

By solving equation (1) & (2)

u = 2.6045 $\frac{km}{hr}$

v = 0.5205 $\frac{km}{hr}$

Now the time taken by the un powered raft to cover this distance

$T = \frac{100}{v}$

Because un powered raft travel with the speed of the current.

$T = \frac{100}{0.5205}$

T = 192.12 hr

Therefore the time taken by the un powered raft to cover this distance is

T = 192.12 hr

3 0

4 years ago

Read 2 more answers

Explain how to multiply the following whole numbers 21 x 14

Lesechka [4]

Answer:

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Step-by-step explanation:

Given

$21\:\times \:14$

Line up the numbers

$\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}$

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

Multiply the bold numbers: 1×4=4

$\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}$

Multiply the bold numbers: 2×4=8

$\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the top number by the bolded digit of the bottom number

$\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the bold numbers: 1×1=1

$\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}$

Multiply the bold numbers: 2×1=2

$\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}$

Add the rows to get the answer. For simplicity, fill in trailing zeros.

$\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}$

adding portion

$\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}$

Add the digits of the right-most column: 4+0=4

$\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}$

Add the digits of the right-most column: 8+1=9

$\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}$

Add the digits of the right-most column: 0+2=2

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Therefore,

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

6 0

3 years ago