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Tomtit [17]
3 years ago
15

Read the excerpt from "An Occurrence at Owl Creek Bridge."

Mathematics
2 answers:
Trava [24]3 years ago
7 0
It represents death and dying as when you die it is said that you see a light.

Hope this helps :)
Sever21 [200]3 years ago
4 0

The answer is

<em>a. death and dying</em>

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Which correlation does the two-way table suggest?
natima [27]

Answer:

The answer is C.

Hope this helps :)

<em>-ilovejiminssi♡</em>

6 0
3 years ago
Find the value of a and b
andriy [413]

Answer:

a = 133 degrees

b = 78 degrees

Step-by-step explanation:

the top and bottom lines are parallel.

the two sidelines are lines that intercept the top and bottom lines.

as they intercept parallel lines, they actually must have the same angles with them.

so, the 47 degrees inner angle at the bottom line, must be also somewhere at the interception point with the top line. and right, it must be now mirrored the outward angle at the top line. and that means a (the inward angle at the top line) is also the outward angle at the bottom line.

the sum of inward and outward angles at a point must always be 180 degrees.

so, the outward angle of 47 = the inward angle a =

= 180 - 47 = 133 degrees.

similar in the other side.

102 is the inward angle.

the outward angle of that is 180 - 102 = 78 degrees.

and that is also the inward angle b.

b = 78 degrees

4 0
3 years ago
A company must select 4 candidates to interview from a list of 12, which consist of 8 men and 4 women.
Radda [10]
Part A

If 4 candidates were to be selected regardless of gender, that means that 4 candidates is to be selected from 12.

The number of possible selections of 4 candidates from 12 is given by

^{12}C_4= \frac{12!}{4!(12-4)!}= \frac{12!}{4!\times8!} =11\times5\times9=495

Therefore, the number of <span>selections of 4 candidates regardless of gender is 495.



Part B:

</span>
<span>If 4 candidates were to be selected such that 2 women must be selected, that means that 2 men candidates is to be selected from 8 and 2 women candidates is to be selected from 4.

The number of possible selections of </span><span>2 men candidates from 8 and 2 women candidates from 4 is given by

</span><span>^{8}C_2\times ^{4}C_2= \frac{8!}{2!(8-2)!}\times &#10;\frac{4!}{2!(4-2)!} \\  \\ = &#10;\frac{8!}{2!\times6!}\times\frac{4!}{2!\times2!} &#10;=4\times7\times2\times3=168

Therefore, the number of selections of 4 candidates </span><span>such that 2 women must be selected is 168.</span>



Part 3:

If 4 candidates were to be selected such that at least 2 women must be selected, that means that 2 men candidates is to be selected from 8 and 2 women candidates is to be selected from 4 or 1 man candidates is to be selected from 8 and 3 women candidates is to be selected from 4 of <span>no man candidates is to be selected from 8 and 4 women candidates is to be selected from 4.

The number of possible selections of </span>2 men candidates from 8 and 2 women candidates from 4 of <span>1 man candidates from 8 and 3 women candidates from 4 of no man candidates from 8 and 4 women candidates from 4 is given by
</span><span>
^{8}C_2\times ^{4}C_2+ ^{8}C_1\times ^{4}C_3+ ^{8}C_0\times ^{4}C_4 \\  \\ = \frac{8!}{2!(8-2)!}\times \frac{4!}{2!(4-2)!}+\frac{8!}{1!(8-1)!}\times \frac{4!}{3!(4-3)!}+\frac{8!}{0!(8-0)!}\times \frac{4!}{4!(4-4)!} \\ \\ = \frac{8!}{2!\times6!}\times\frac{4!}{2!\times2!}+\frac{8!}{1!\times7!}\times\frac{4!}{3!\times1!}+\frac{8!}{0!\times8!}\times\frac{4!}{4!\times0!} \\  \\  =4\times7\times2\times3+8\times4+1\times1=168+32+1=201

Therefore, the number of selections of 4 candidates </span><span>such that at least 2 women must be selected is 201.</span>

7 0
3 years ago
Read 2 more answers
43 milk jugs are in a fridge with 11 that are spoiled. 5 jugs are randomly selected. what are the odds that the first two are go
malfutka [58]

Answer:

0.85% probability that the first two are good and the last three are spoiled

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the first two jugs is not important, as is not the order in which the last three are selected. So we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Desired outcomes:

2 spoiled, from a set of 43-11 = 32.

3 spoiled, from a set of 11.

So

D = C_{32,2}*C_{11,3} = \frac{32!}{2!(32-2)!}*\frac{11!}{3!(11-3)!} = 982080

Total outcomes:

5 jugs from a set of 43.

T = C_{43,5} = \frac{43!}{5!(43-5)!} = 115511760

Probability:

p = \frac{D}{T} = \frac{982080}{115511760} = 0.0085

0.85% probability that the first two are good and the last three are spoiled

4 0
3 years ago
Y=x+2 x input -2 -1 0 1 2​
Alex_Xolod [135]
As you plug in those numbers for x (from left to right) y=
0
1
2
3
4
8 0
3 years ago
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