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Liono4ka [1.6K]
3 years ago
11

Which value is a simplified form of 3^5 / 3^-5 ?

Mathematics
1 answer:
vazorg [7]3 years ago
6 0

Answer:

The answer is 3¹⁰

Step-by-step explanation:

hope this helps

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Which of those is an example of numerical data ? A. Average speed b. Type of pet C. Favorite car d.home state
garri49 [273]
Average speed the others are categorical.
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What’s the slope?? Please help!
9966 [12]

Answer:

I believe the slope of the line is 5

Step-by-step explanation:

5 + 5 = 10

10 + 5 + 15

15 + 5 = 20

Hope this helps :)

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3 years ago
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6 girls share 4 brownies equally. What fraction of the brownies does each friend get?
aliina [53]

Answer:

2/3

Step-by-step explanation:

You have 6 friends who want to share 4 brownies equally. You simply divide the 4 brownies by 6 people:

4/6 = 2/3 (simplified)

So each friend gets 2/3 of a brownies

[RevyBreeze]

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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. 
<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
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A person is studying the relationship between temperature and mental ability. So,
valina [46]

Answer:

control

Step-by-step explanation:

3 0
3 years ago
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