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3 years ago

The food stand at the zoo sold 2514

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

4 0

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Well's College in New York was previously an all girls school. In 2005, the college began to allow boys to enroll. By 2012, the

kogti [31]

Answer: 150 boys

Step-by-step explanation:

Let the number of boys be x

Since there were 200 more girls than boys, the number of girls will be: = x + 200

Since the ratio of boys to girls was 3:7, this can be solved further below:

3/7 = x/x+200

Cross multiply

(7 × x) = 3(x + 200)

7x = 3x + 600

7x - 3x = 600

4x = 600

x = 600/4

x = 150

The number of boys enrolled is 150

Number of Girls will be:

= X + 200

= 150 + 200

= 350

Check: Number of boys / Number of girls

= 150/350

= 3/7

= 3:7

7 0

2 years ago

Anyone know how to do this??!?!

ludmilkaskok [199]

Add all the numbers of the triangle and then divide them by the sides

7 0

3 years ago

Please help with this!!

lozanna [386]

Answer:

#1.---60% #2.---$12.60 per hour

Step-by-step explanation:

#1. Since It was originally $50 you have 50% already, so what I did was I multiplied 50 by 2. this got me 100 which is a simpler way to calculate by. 50 to 80 dollars is a $30 change, so multiply that by 2 and you get 60. 60 will be your percent of change. You can double check this by multiplying 50 by .60, the answer will be 30 and 50+30=80

#2 Since Tyler starts off with $12 per hour with a 5% increase you will need to multiply 12*.05 this will get you the answer of 0.6 you then need to add $12, since that was his original payment. 12+0.6=12.60, so his final payment would by $12.60 per hour

5 0

2 years ago

A fishing lake was stocked with 350 trout. each year,the population decreases by 14. the population of trout in the lake after x

Lubov Fominskaja [6]

Check the attached file for the solution.

7 0

3 years ago

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