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Black_prince [1.1K]
3 years ago
11

Billy watched the clown car drive up to the circus and saw 14 clowns get out of one little car. The clowns made up 56%, percent

of the performers in the circus. How many performers were in the circus?
Mathematics
2 answers:
Mumz [18]3 years ago
4 0
0.56x = 14 

<span>x = 14 / 0.56 </span>

<span>x = 25 performers</span>
KATRIN_1 [288]3 years ago
4 0
You would solve using a proportion. 14/.56 = x/.56 (x is the unknown). Cross multiply and solve for x. Your answer is 25 total performers.
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596 divided by 3 using long divison
Basile [38]

Answer:

198.6 with the remaining 6

Step-by-step explanation:

596/3

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596/3 = 198.67 to the nearest hundredth

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3 years ago
Find the midpoint of R(2,2) and S(-3,6)
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Step-by-step explanation:

Midpoint formula: (\frac{x1+x2}{2} ,\frac{y1+y2}{2} )

(x1, x2)=(2, -3)

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 (\frac{x1+x2}{2} ,\frac{y1+y2}{2} )

=(\frac{2-3}{2} ,\frac{2+6}{2} )

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Hope this helps!! :)

Please let me know if you have any question or need further explanation

3 0
3 years ago
Read 2 more answers
Amanda earned a score of 940 on a national achievement test that was normally distributed. The mean test score was 850 with a st
hichkok12 [17]

Answer:

lower than Amanda:  816 students

Step-by-step explanation:

An equivalent way in which to state this problem is:  Find the area under the standard normal curve to the left (below) 940.

Most modern calculators have built in distribution functions.

In this case I entered the single command   normalcdf(-1000,940, 850, 100)

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In this particular situation, this means that 0.816(1000 students) scored lower than Amanda:  816 students.

6 0
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<span>Let C denote the number of candidates they interview and E the number of employees they train.
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If <span>it takes 120 hours and $3600 to train an employee, then it takes 120E hours and $3600E to train E employees.
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Company has less than <span>$95000, then 400C+3600E<95000.
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Company <span>wants to spend at most 470 hours, then 20C+120E\le&#10; 470.
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<span>You obtain the system of two inequalities:
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\left \{ {{400C+3600E\ \textless \ 95000} \atop {20C+120E\le&#10; 470}} \right.  \\  \left \{ {{4C+36E\ \textless \ 950} \atop {C+6E\le &#10;23.5}} \right. \\ \left \{ {{4C+36E\ \textless \ 950} \atop {4C+24E\le &#10;94}} \right.
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3 0
3 years ago
Need help please now
ZanzabumX [31]
It’s A letter A hope this helped
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3 years ago
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