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ICE Princess25 [194]
2 years ago
11

The yearly cost in dollars, y, at a video game arcade based on total game tokens purchased, x, is y = y equals StartFraction 1 O

ver 10 EndFraction x plus 60.x + 60 for a member and y = y equals StartFraction 1 Over 5 EndFraction x. x for a nonmember. Explain how the graph of a nonmember’s yearly cost will differ from the graph of a member’s yearly cost.
Mathematics
2 answers:
quester [9]2 years ago
8 0

Answer:

The graph of a nonmember’s yearly cost will be steeper, but start lower than the graph of a member’s yearly cost.

Step-by-step explanation:

The graph of a nonmember’s yearly cost will be steeper, but start lower than the graph of a member’s yearly cost.

Guest
1 year ago
this is the sample yall stay safe
klio [65]2 years ago
6 0

Answer:

The members slope is flatter by half compared to the non-members slope. The graphs intersect at 600 tokens and $120. Until that point the cost for non-members is less than the cost for members. After 600 tokens the cost for members is less than the cost for non-members.

Step-by-step explanation:

I hope I have your equations correct:

Member:

y = 1/10x + 60

Non-Member:

y = 1/5x

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Answer:

Therefore,

There are Four i.e 4 terms in the expression,

-7+12x^{4}-5y^{8}+x

Step-by-step explanation:

Algebraic Expression:

An algebraic expression is a mathematical expression that consists of variables, numbers and operations.

  • Variables with the coefficient is called as Term.
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-7+12x^{4}-5y^{8}+x

So,

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Therefore,

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-7+12x^{4}-5y^{8}+x

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A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d
tankabanditka [31]
The selection of r objects out of n is done in

C(n, r)= \frac{n!}{r!(n-r)!} many ways.

The total number of selections 10 that we can make from 6+7=13 students is 

C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}=  286
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A. 
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.


</span>C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35
<span>
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B.
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with the same logic as in A, the number of groups were all 7 boys are in, is 

</span>C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20
<span>
so the probability is 20/286=0.07

C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls. 

</span>C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105
<span>
the probability is 105/286=0.367

since  case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"

selecting 5 boys and 5 girls can be done in 

</span>C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126

many ways,

so the probability is 126/286=0.44
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