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2 years ago

May someone help me please?

Mathematics

1 answer:

Anastaziya [24]2 years ago

5 0

Since we know that g(x) is between h(x) and f(x) we can use the same formula format.
If g(x) = a * b^x
a = 160 most likely
b^x would be between 4/5 and 1/5
Look at desmos example below where the third is 3/5^x and lies directly between the first 2

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In basketball, some baskets are worth two points. Others are worth three points. In one game, the ratio of three-point basket

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Answer:

Writing the information in the form of ratio

Three point basket = Three point tries

3 : 4

27 : x

Think about multiplication , how many 3s do you need to make 27 ?

Its 27 ÷ 3= 9

Because you need 9 lots of 3, Then to make it proportional , you'd need 9 lots of 4

9×4 = 36

Three points tries is 36

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2 years ago

The perimeter of a regular pentagon is 45 inches. Find the length of each side

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Answer: 9

Step-by-step explanation: pentagon has 5 sides so divide 45 by 5

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2 years ago

Please help I don't understand!

Dafna11 [192]

Answer:

Attach the image of the triangle so I can help out and edit my answer

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1 year ago

11 more than x is less than. 3 (please)

KATRIN_1 [288]

Answer:

(11+x)-3

Step-by-step explanation:

11 more than x is another way of saying 11+x

And "less than 3" means subtracting 3 from the terms

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(11+x) -3

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2 years ago

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.
"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.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

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

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

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

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

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

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.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

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.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

6 0

3 years ago

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