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3 years ago

10

A small Learning Center holds twenty-four less students than a large Learning Center. Together the Learning Centers hold seventy

-two students. How many students does each Learning Center hold?

1 answer:

olga2289 [7]3 years ago

6 0

Answer:

A small learning center holds 48 students

Step-by-step explanation:

A small Learning Center holds <em><u>twenty-four less</u></em> students <u><em>than a large Learning Center.</em></u> Together the <em><u>Learning Centers hold seventy-two</u></em> students. How many students does each Learning Center hold?

72-24=48

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PilotLPTM [1.2K]

Answer:

$79.20

Step-by-step explanation:

fixed charged of $19 x 4 months = $76.00

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Therefore, $76.00 + $3.20 = $79.20

8 0

3 years ago

What simple interest rate would allow $6000 to grow to an amount of $14550 in 10 years?

Harman [31]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{ \: 14.25 \: \% \: }}}}}$

Step-by-step explanation:

Given,

Principal ( P ) = $ 6000

Amount ( A ) = $ 14550

Time ( T ) = 10 years

Rate ( R ) = ?

<u>Finding </u><u>the </u><u>Interest</u>

The sum of principal and interest is called an amount.

From the definition,

$\boxed{ \sf{Amount = \: Principal + Interest}}$

plug the values

⇒ $\sf{14550 = 6000 + Interest}$

Swap the sides of the equation

⇒ $\sf{6000 + Interest = 14550}$

Move 6000 to right hand side and change its sign

⇒ $\sf{Interest = 14550 - 6000}$

Subtract 6000 from 14550

⇒ $\sf{Interest = \: 8550 \: }$

Interest = $ 8550

<u>Finding </u><u>the </u><u>rate </u>

${ \boxed{ \sf{Rate = \frac{Interest \times 100}{Principal \times Time}}}}$

plug the values

⇒ $\sf{ Rate = \frac{8550 \times 100}{6000 \times 10} }$

Calculate

⇒ $\sf{Rate = \frac{855000}{60000} }$

⇒ $\sf{Rate = 14.25 \: \% \: }$

Hope I helped!

Best regards!!

8 0

3 years ago

5y+5x+3=2 solve for y

denpristay [2]

Answer:y=-1/5-x

Step-by-step explanation:

3 0

3 years ago

Here are two sets of numbers, a and b. Set a :200,104,100,160. Set b: 270,400,483,300, x. Mean of set a: mean of set b= 3:8. Wor

LenKa [72]

Given:

The given sets are:

Set a : 200, 104, 100, 160.

Set b: 270, 400, 483, 300, x.

Mean of set a: mean of set b= 3:8

To find:

The value of x.

Solution:

Formula for mean:

$Mean=\dfrac{\text{Sum of observations}}{\text{Number of observation}}$

The mean of set of a is:

$Mean=\dfrac{200+104+100+160}{4}$

$Mean=\dfrac{564}{4}$

$Mean=141$

The mean of set of b is:

$Mean=\dfrac{270+400+483+300+x}{5}$

$Mean=\dfrac{1453+x}{5}$

$Mean=\dfrac{1453+x}{5}$

It is given that,

Mean of set a: mean of set b= 3:8

$\dfrac{141}{\dfrac{1453+x}{5}}=\dfrac{3}{8}$

$\dfrac{705}{1453+x}=\dfrac{3}{8}$

$8\times 705=3\times (1453+x)$

$5640=4359
+3x$

Isolate the variable x.

$5640-4359
=3x$

$1281
=3x$

Divide both sides by 3.

$\dfrac{1281}{3}
=x$

$427
=x$

Therefore, the value of x is 427.

3 0

3 years ago

The football team has a total of 50 jerseys. There are 35 medium-sized jerseys. What percent of the jerseys are medium-sized j

Mariulka [41]

Hello,

This problem can be simply solved by putting everything into fractions. Since there's a total of 50 shirts but 35 of them are medium sized then the fraction would be 35/50. You're trying to find the percentage so the 2nd fraction would be out of 100 so it would be ?/100.
35 ?
---- = ----
50 100
Now you diagonal multiplication then downwards division.
100 x 35=3500
3500/50=70
So your answer is 70%!! :D
Hope this helps a bunch!!!

4 0

4 years ago