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lara31 [8.8K]
3 years ago
11

What is true about the dilation?

Mathematics
2 answers:
Svet_ta [14]3 years ago
6 0

Answer:

<h2>The answer is D.</h2>

Step-by-step explanation:

This is because only a reduction can be 0-1, as it will shrink, while if you multiply a number by a number greater than 1, is will largen.

OlgaM077 [116]3 years ago
5 0

Answer:

<h2>pretty sure its A sorry if its wrong  </h2>

Step-by-step explanation:

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Factorise: x^2/y^2+1+y^2/X^2​
rewona [7]

Answer:

<h2>( \frac{x}{y}  +  \frac{y}{x}  + 1)( \frac{x}{y}  +  \frac{y}{x}  - 1)</h2>

Step-by-step explanation:

Hope it is helpful.....

7 0
3 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
Read 2 more answers
The Thompson family is buying a car that can travel 70 miles
gogolik [260]

Answer:

The car gets 35 miles per gallons.

Step-by-step explanation:

4 0
3 years ago
School a rented 1 van and 3 buses with 89 students. School b rented 4vans and 3 buses with. 131 students. Each van and each bus
iogann1982 [59]

Answer:18

Step-by-step explanation:

3 0
3 years ago
Financial mathematics (see picture)
Paraphin [41]

Answer:

i think it's the answer in the pic I have send you okk see and mark me brainliest answer okk thanks

6 0
3 years ago
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