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3 years ago

What is the awnser to this question because I don't understand

Mathematics

1 answer:

Virty [35]3 years ago

4 0

Pemdas 1. parentheses: everything is already in parentheses. so next is exponents. 10×-2 is -20. and 10×6 is 60. so 4×-20 is the next step. it equals -80. and then 3×60 is 180. lastly,add -80 and 180 together. the answer should be 100 if i did that correctly.

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PLEASE ANSWER ACCURATELY! TYSM!

murzikaleks [220]

Answer:

Step-by-step explanation:

If you look at your y value which would be your Output you can see the number 4 repeated multiple times

Therefore leading up to the answer that there is more than what output.

The input is the number you feed into the expression, and the output is what you get after the look-up work or calculations are finished.

Hope this helps :)

6 0

2 years ago

Which of the following expressions are equivalent to 18x + 12 – 4? Select three that apply.

GalinKa [24]

Answer:

C 18x + 8

D 2(9x + 4)

F 6(3x + 2) – 4

Step-by-step explanation:

hope this helped you =)

5 0

3 years ago

Read 2 more answers

Please help! / 50 points

lutik1710 [3]

Answer:

Solution given:

radius [r]=4

height [h]=5

we have

Surface area of cylinder=2πr(r+h)

=2π×4(4+5)

=72πunits³ is a required volume.

3 0

3 years ago

The area is 851 square meters if the length is 23 meters what is the width

jarptica [38.1K]

851÷23= 037 (hope this helps) 69 161 161 0

6 0

3 years ago

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