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3 years ago

Hey I need this answered fast!! and if u please could help me I have more questions similar to this one on my page

Mathematics

1 answer:

Tatiana [17]3 years ago

4 0

Answer:

B and E

Step-by-step explanation:

The zero of a function is whatever point on the function where the y-value is zero. Hope this helped.

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Expanded Notation Form:
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3 years ago

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3 years ago

135÷5, draw two different ways to break apart the array . use the distributive property to write products for each different way

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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

3 years ago