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maks197457 [2]
3 years ago
14

Systems By Graphing - Solve All Correctly For a Thanks, 5-Stars, And A Brianliest!

Mathematics
1 answer:
BartSMP [9]3 years ago
4 0

Answer:

Step-by-step explanation:

Amari= Graph C and Solution is (-6,-2)

Bella= Graph A and Solution is (3,4)

Carl= Graph B and Solution is (0,-3)

0 solutions because the slopes are the same. The lines will never cross because they are at the exact same angle.

No, I do not agree with Joey because the lines have different slopes and will lead to the system cross which is the solution.

1st graph: y=2x-1 and y=7 solution #1= (4,7)

2nd graph: y=-2x-3 and y=1/2x solution #2= (-2,1)

3rd graph: y=x and y= -1/5+6 solution #3= (5,5)

I had this exact same assignment a few months ago, my teacher didn't use the 2nd slide but I had the 1st and 3rd slide so this should help!

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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
thus, the sample space of the experiment is 286

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>
P(all 6 girls chosen)=35/286=0.12

B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"

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
6 0
3 years ago
Read 2 more answers
Question 24 of 25 Use the quadratic formula to find the solutions to the equation. x² – 3x+1=0
user100 [1]
I think the answer is B! Hope this helps!
3 0
3 years ago
Plz help with this problem!<br><br> x^4 + x^2 + 1
TEA [102]

Answer:

You cannot simplify this problem down anymore; it is at its simplest form.

Step-by-step explanation:

You cannot add together x^4 and x^2 due to the simple fact that they do not share similar exponents.

For example you could add together x^4 and x^4 to get 2x^4. But you cannot add x^4 and x^2.

7 0
3 years ago
Seventy-eight is 26% of what number?
tensa zangetsu [6.8K]
78 / 0.26 = 300 so the answer is 300
8 0
3 years ago
Chen chooses a prime number he multiple s it by 10 and then rounds it to the nearest hundred his answer is 400 write all the pos
poizon [28]
It would most likely be 41 because when you would round 41 after you would × 41 times 10 equals 410 then you would round it to 400
6 0
3 years ago
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