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3 years ago

Explain answer !! While you answer

Mathematics

1 answer:

Iteru [2.4K]3 years ago

7 0

Answer:

it's c

Step-by-step explanation:

here's some proof bud thanks

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Which is the graph x>2?

LenaWriter [7]

Answer:

option C will be the correct answer here

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3 years ago

What is the answer!!! 35.00 for a jacket. 10% discount 5% sales tax whats the final cost? PLEASE HELP ME

ASHA 777 [7]

Firstly, you should find the discount that is taken off of the jacket, and then also find the sales tax.
As the discount is 10%, you simply have to divide 35 by 10. 35/10= 3.50.
Therefore, you need to subtract the 3.50 from the 35, which gives you 31.50. You've then got the 5% sales tax to add on, which is 3.50/2= 1.75, you've then got to add 1.75 to 31.50, which gives you 33.25.
However, if the sales tax is on the discounted price, then it will be 31.50/10= 3.15/2= 1.58, and you've then got to add this is to discount price, which is 31.50+1.58= 33.08

Hope this helps :)

6 0

3 years ago

What are the coordinates of C if translated 5 units upward?

aleksklad [387]

(6, 11)

hope this helps!

4 0

3 years ago

7 less than three fifths of b is a

Citrus2011 [14]

It would be 3/5b-7=y?

4 0

3 years ago

Read 2 more answers

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

4 years ago