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4 years ago

Which of the following is an equivalent expression?

Mathematics

1 answer:

Marina86 [1]4 years ago

6 0

I'm pretty sure that the answer is B.

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Which products are greater than 2 5 6 ? Select all that apply.

Lelechka [254]

Answer:

Step-by-step explanation:

6 it is 6 I hope this helps

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3 years ago

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Find the midpoint of the segment with the following endpoints. (8,1) and (5,7)

Crazy boy [7]

The midpoint would be (13/2 , 4)

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3 years ago

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HELP I’M GONNA FAIL!!!!!! Pls help ASAP! I NEED TO PROMOTEEEEEE!!!!

Nana76 [90]

Volume of sphere is
V=4/3pi r^3

Where r is 2 in this example so,

V= 10.67 pi
V= 33.51

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2 years ago

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In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, u

horsena [70]

Answer:

$\large\boxed{\large\boxed{0.119}}$

Explanation:

You need to find the probability that exactly three of the first 11 inspected packages are damaged and the fourth is damaged too.

1. Start with the first 11 inspected packages:

a) The number of combinations in which 11 packages can be taken from the 20 available packages is given by the combinatory formula:

$C(m,n)=\dfrac{m!}{m!(m-n)!}$

$C(20,11)=\dfrac{20!}{11!\cdot(20-11)!}$

b) The number of combinations in which 3 damaged packages can be chossen from 7 damaged packages is:

$C(7,3)=\dfrac{7!}{3!\cdot(7-3)!}$

c) The number of cominations in which 8 good packages can be choosen from 13 good pacakes is:

$C(13,8)=\dfrac{13!}{8!\cdot(13-8)!}$

d) The number of cominations in which 3 damaged packages and 8 good packages are chosen in the first 11 selections is:

$C(7,3)\times C(13,8)$

e) The probability is the number of favorable outcomes divided by the number of possible outcomes, then that is:

$\dfrac{C(7,3)\times C(13,8)}{C(20,11)}$

Subsituting:

$\dfrac{\dfrac{7!}{3!\cdot(7-3)!}\times \dfrac{13!}{8!\cdot(13-8)!}}{\dfrac{20!}{11!\cdot(20-11)!}}$

$=\dfrac{\dfrac{7!}{3!\cdot 4!}\times \dfrac{13!}{8!\cdot 5!}}{\dfrac{20!}{11!\cdot 9!}}=0.26818885$

2. The 12th package

The probability 12th package is damaged too is 7 - 3 = 4, out of 20 - 11 = 9:

3. Finally

The probability that exactly 12 packages are inspected to find exactly 4 damaged packages is the product of the two calculated probabilities:

$0.26818885\times 4/9=0.119$

6 0

3 years ago

What is the answer?.....

Butoxors [25]

Answer:

Step-by-step explanation:

f(2) = 2^3 - 4(2)^2 = 8 - 4(4) = 8 - 16 = -8

g(-8)= 4(-8) + 5 = -32 + 5 = -27

C is the answer

8 0

3 years ago