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kozerog [31]
3 years ago
14

Deyonne made 25% more cookies than Cameron. They sold cookies for $0.50 each. Let x represent the number of cookies Cameron made

. Which of the following expressions represent possible revenue from the sales. Select all that apply.
A. 0.50[x + (x + 0.25x)]
B. 0.50(x + 0.25x)
C. 0.50(x + 1.25x)
D. 0.50(0.75x + x)
E. 0.50(1.75x)
F. 0.50(2.25x
Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
5 0
First we define the varible to use:
 x = represent the number of cookies
 Cameron:
 
The amount of cookies he made was:
 x
 Deyonne:
 
We can rewrite the amount in different ways:
 x + 0.25x
 1.25x
 Then, the total revenue is:
 0.50 * (x + (x + 0.25x))
 We rewrite:
 0.50 * (x + 1.25x)
 0.50 * (2.25x)
 Answer:
 
The following expressions represent possible revenue from the sales:
 
F. 0.50 (2.25x
 
C. 0.50 (x + 1.25x)
 
A. 0.50 [x + (x + 0.25x)]
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marishachu [46]

Answer:

a) 286

b) 1,037,836,800

c) 285

Step-by-step explanation:

a) Since order is not important, the total possible number of ways to choose 10 players out of 13 is the following combination:

C(13,10)=\frac{13!}{(13-10)!10! } \\C(13,10)=\frac{13*12*11*10!}{(3*2*1)10! }\\C(13,10) = 286

b) The total number of possibilities to assign positions by the selecting 10 players is the permutation of 13 players for 10 positions:

P(13,10) = \frac{13!}{(13-10)!}\\P(13,10) = 13*12*11*10*9*8*7*6*5*4\\P(13,10) =1,037,836,800

c) The number of ways to pick 10 players including at least one woman is equal to the total number of ways to pick 10 players (found in item a) minus the the number of ways to pick 10 players without picking a single woman.

Since there 10 male players for 10 positions, there is only one possible way to pick a team without women, therefore:

P=286-1 =285

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If Kiontre reads 100 pages in 5 days, how many pages does he read in 1 day?
Aleks04 [339]

Answer: 20 pages

Step-by-step explanation: 100/5 is 20

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Lori creates the following design for a T-shirt.
kakasveta [241]

<u>Solution-</u>

From the figure,

AE = 2.4

EB = 2.8

BC = 11.7


Area of rectangle 1 = 8.68 sq.in

\Rightarrow FH \times HI=8.68

\Rightarrow EB \times HI=8.68  (∵ sides of the rectangle 2)

\Rightarrow 2.8 \times HI=8.68

\Rightarrow HI=3.1


Area of Triangle 1 = 6.48 sq.in

\Rightarrow \frac{1}{2}\times AE \times EG= 6.48

\Rightarrow \frac{1}{2}\times AE \times (EF+FG)= 6.48

\Rightarrow \frac{1}{2}\times AE \times (EF+HI)= 6.48  (∵ sides of the rectangle 1)

\Rightarrow EF+3.1= 5.4

\Rightarrow EF=2.3

\Rightarrow BH=2.3  (∵ sides of the rectangle 2)


BC = BH+HI+IC

\Rightarrow 11.7= 2.3+3.1+IC

\Rightarrow IC=6.3


The area of Rectangle 2,

=EB\times BH =2.8\times 2.3=6.44\ sq.in


The area of Triangle 2,

\frac{1}{2}\times GI \times IC=\frac{1}{2}\times EB \times IC=\frac{1}{2}\times 2.8 \times 6.3=8.82\ sq.in


The area of the whole figure = Area of Triangle 1 + Area of rectangle 1 + Area of Triangle 2 + Area of rectangle 2

= 6.48+8.68+8.82+6.44=30.42 sq.in


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