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Sergio [31]
3 years ago
7

student enrollment at a local school is concerning the community because the number of student has dropped to 504 which is a 20%

decease form the previous year. what was the student enrollment the previous year?
Mathematics
1 answer:
zubka84 [21]3 years ago
4 0
Since 504 is equal to 80% od the previous years enrollment you would divide 504 by 80 which gives you 6.3 which is 1% of the previous years enrollment.  You would then multiply 6.3 by 100 in order to find 100% of the previous years enrollment.  to check your work divide 630 by 20% or .20 which will give you 126. You then subtract 126 from 630 which gives you 504.
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Arithmetic and Geometric Sequences (Context)
V125BC [204]

The formula for compound interest

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

A is the amount in the account at the end

P is the principal balance or the amount initially invested

r is the annual interest rate in decimal form

n is the number of times it is coupounded per year

t is the number of years

A = 1800 ( 1+ .0375/1) ^ (1*6)

A = 1800 ( 1.0375)^6

A = 2244.92138

Rounding to the nearest cent

A = 2244.92

7 0
1 year ago
The perimeter of the square below is 36 yards. What is the length of each side? <br> yards
Inessa05 [86]

Answer:

9 yards

Step-by-step explanation:

36 divided by 4 = 9

3 0
2 years ago
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Anna11 [10]
[ ( x + 4 )( x + 5 ) + 4( x + 1 )( x +  5) - 5( x + 1 )( x + 4 ) ] / [( x + 1 )( x +  4)( x + 5 )] = ( x^2 + 9x + 20 + 4x^2 + 24x +20 - 5x^2 - 25x - 20) / [( x + 1 )( x +  4)( x + 5 )] =
( - 2x + 20 ) /  [( x + 1 )( x +  4)( x + 5 )] = ( - 2)( x - 10) / [( x + 1 )( x +  4)( x + 5 )]
4 0
3 years ago
The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding
melamori03 [73]

Answer:

We conclude that the mean wedding cost is less than $30,000 as advertised.

Step-by-step explanation:

We are given the following data set:(in thousands)

29100, 28500, 28800, 29400, 29800, 29800, 30100, 30600

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{236100}{8} = 29512.5

Sum of squares of differences = 3408750

S.D = \sqrt{\frac{3408750}{7}} = 697.82

Population mean, μ = $30,000

Sample mean, \bar{x} = $29512.5

Sample size, n = 8

Alpha, α = 0.05

Sample standard deviation, s = $ 697.82

First, we design the null and the alternate hypothesis

H_{0}: \mu = 30000\text{ dollars}\\H_A: \mu < 30000\text{ dollars} We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{29512.5 - 30000}{\frac{697.82}{\sqrt{8}} } = -1.975

Now,

t_{critical} \text{ at 0.05 level of significance, 7 degree of freedom } = -1.894

Since,                  

t_{stat} < t_{critical}

We fail to accept the null hypothesis and reject it.

We conclude that the mean wedding cost is less than $30,000 as advertised.

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Answer:

There are only 4 operations. Joanna is incorrect and Claudia is correct.

Hope this helps :)

-jp524

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2 years ago
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