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Nataliya [291]
2 years ago
14

If f(x) = -8 - 5x, what is f(-4)

Mathematics
2 answers:
Mashcka [7]2 years ago
6 0

Answer:

f(- 4) = 12

Step-by-step explanation:

To evaluate f(- 4) substitute x = - 4 into f(x) , that is

f(- 4) = - 8 - 5(- 4) = - 8 + 20 = 12

bogdanovich [222]2 years ago
3 0

f(x)=-8-5x

\\ \rm\longmapsto f(-4)

  • Put -4 inplace of x

\\ \rm\longmapsto -8-5(-4)

\\ \rm\longmapsto -8+20

\\ \rm\longmapsto 12

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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
What is the value of x after these statements if the starting value of x is 3.
kolbaska11 [484]
[A]
3^{2}=9
5^{3}=125
9\neq25

Ignore


[B]
3^{2} = 9
9 \ \textgreater \  3
Yes

3^{3}=27
4( 3^{2})=36&#10;&#10;27 \ \textless \  36
Yes

Therefore
x=3+2=5


[C]
4^{5}=1024&#10;&#10; 5^{4}=625&#10;&#10;1024\ \textless \ 625
No

5^{5}=3125&#10;&#10; 5^{5}=3125&#10;&#10;3125\ \textgreater \ 3125
No

Ignore


[D]
5^{5}=3125&#10;&#10; 2^{5}=32&#10;&#10;3125\ \textgreater \ 32
Yes

5^{2}=25&#10;25\ \textless \ 11
No

Therefore
x=5+8=13


[E]
8\ \textgreater \ 7
Yes

Therefore
x = 13^{2}=169
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3 years ago
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schepotkina [342]
V=763.02
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5 0
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Answer:

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Step-by-step explanation:

"sum" tells you to add the two numbers together.

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