456 students participated in switch day the students raised money for charity so that the principal would approve of the day if the total amount of money raised was $912 and each student brought in the same amount of money how much did each student raise?

Answers

Answer 1

Answer: Well according to my calculations, each student brought in $2

Hope that helps :D

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The difference of x and a number is 6. What is the other number?A) 6-x
B) x-6
C) 6-(x-6)
D) 6+x

Answers

The answer would be B because difference means subtracting so the answer would be x-6

I really need help!! this is for the end of year assessment and i have to get this last one right so i don't have an F in math.Sierra and Alek each have five cousins. Their ages are listed below.

Sierra’s cousins: 2, 11, 12, 13, 15
Alek’s cousins: 9, 11, 11, 12, 13

Which of the following statements is true?
A.For both sets of data, the median is equal to the mean.
B.The ages of Sierra's cousins are more spread out than those of Alek's cousins.
C.The mean age of Sierra's cousins is greater than that of Alek's cousins.
D.There are no outliers.

Answers

Answer: B: The ages of Sierra's cousins are more spread out than those of Alek's cousin.

john invested $2000 that earn interest at 4% per annum compounded monthly. two year later the interest rate is 4.50% compounded quaterly. determine the accumulated value of investment two year after the change

Answers

Answer:

$\$2,369.11$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+(r)/(n))^(nt)$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

Part 1)

$t=2\ years\n P=\$2,000\n r=0.04\nn=12$

substitute in the formula above

$A=2,000(1+(0.04)/(12))^(12*2)$

$A=2,000(1.0033)^(24)$

$A=\$2,166.29$

Part 2)

$t=2\ years\n P=\$2,166.29\n r=0.045\nn=4$

substitute in the formula above

$A=2,166.29(1+(0.045)/(4))^(4*2)$

$A=2,166.29(1.01125)^(8)$

$A=\$2,369.11$

Between what pair of numbers is the product of 289 and 7

Answers

The product of 289 and 7 is somewhere between 289 and 2,100 .

PLEASE DON'T SKIP

Solve the system of equations.

Answers

      2x + 3y = 1

      y = 3x + 15

There's not much you can do with the first equation, because it has
two variables in it ... 'x' and 'y' . No matter how much you move them
around, you'll never be able to get either one equal to just a number.
Is there any way you could get rid of one of the variables in the first
equation, and have just 1 letter in it to solve for ?

Absolutely ! The second equation tells you something that 'y' is equal to,
(3x + 15). "EQUAL" is very powerful. It means that wherever you see 'y',
you can put (3x + 15) in its place, and you won't change anything or
upset anything. One thing you can do is take that (3x + 15) from the
2nd equation, and put it right into the first equation in place of 'y'.
You'll see how that helps as soon as you do it.

   First equation: 2x + 3y = 1

   Substitute for 'y' : 2x + 3(3x + 15) = 1

Remove parentheses: 2x + 3(3x) + 3(15) = 1
      2x + 9x + 45 = 1

Combine the terms with 'x' in them: 11x + 45 = 1

Look what you have now ! An equation with only one variable in it !

Subtract 45 from each side: 11x = -44

Divide each side by 11 : x = -4

You're more than halfway there. Now you know what 'x' is,
and you can use it with either equation to find what 'y' is.

-- If you use it with the first equation: 2x + 3y = 1

   Put in the value of 'x': 2(-4) + 3y = 1

Remove the parentheses: -8 + 3y = 1

Add 8 to each side:    3y = 9

    Divide each side by 3 : y = 3

-- If you use it with the 2nd equation: y = 3x + 15

      Put in the value of 'x' :     y = 3(-4) + 15

   Remove the parentheses: y = -12 + 15

Add numbers on the right side:   y = 3 (same as the other way)

So there's your solution for the system of two equations:

   x = -4
   y = 3

165°
145
x =
Please Help!!!!