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katen-ka-za [31]
3 years ago
8

What is the domain and range of the graph below?

Mathematics
1 answer:
liubo4ka [24]3 years ago
5 0

9514 1404 393

Answer:

  • domain: all real numbers
  • range: y ≥ 0

Step-by-step explanation:

The domain is the horizontal extent of the graph. For a graph like this, we assume the ends extend to infinity, both horizontally and vertically. Then the horizontal extent (domain) is from -infinity to +infinity: all real numbers.

The graph does not go below y=0, so the vertical extent (range) is y ≥ 0.

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In an anthropology class there are 77 anthropology majors and 1414 non-anthropology majors. 55 students are randomly selected to
Nata [24]

Answer:

P = 0.9989

Step-by-step explanation:

In order to do this, I will use the following numbers to make the calculations easier. In this case, We'll say that we have 7 majors and 14 non majors anthopology to present a topic.

This means that in the class we have 21 students.

Now, we choose 5 of them, and we want to know the probability that 1 of them is non major.

First, we need to calculate the number of ways we can select the students in all cases, and then, the probability.

First, we'll use the combination formula, to calculate the number of ways we can select the 5 students out of the 21. We use combination, because it does not matter the order that the students are selected.

C = m! / n!(m - n)!

Where:

m: number of students

n: number of selected students out of m.

With this expression we will calculate first, how many ways we can choose the 5 students out of 21:

C1 = 21! / 5!(21-5)! = 20,349

Now let's calculate the number of ways you can get the all 5 students are non majors:

C2 = 14! / 5!(14 - 5)! = 2002

Now we need to know the number of ways we can get 4 non majors and 1 major:

C3 = C3' * C3''

C3' represents the number of ways we can get 4 non majors and the C3'' represents the number of ways we can get 1 major.

C3' = 14! / 4!(14 - 4)! = 1,001

C3'' = 7! / 1!(7 - 1)! = 7

C3 = 1001 * 7 = 7,007 ways to get 4 non majors and 1 major

Now the way to get 3 non majors and 2 majors, we do the same thing we do to get 4 non majors and 1 major, but changing the numbers. Then the way to get 2 non majors and 3 majors, and finally 1 non major and 4 majors:

3 non majors and 2 majors:

C4 = C4' * C4'' = [14! / 3!(14 - 3)!] * [7! / 2!(7 - 2)!] = 7,644

2 non majors and 3 majors:

C5 = C5' * C5'' = [14! / 2!(14 - 2)!] * [7! / 3!(7 - 3)!] = 3,185

1 non major and 4 majors:

C6 = C6' * C6'' = [14! / 1!(14 - 1)!] * [7! / 4!(7 - 4)!] = 490

Finally to know the probability of getting 1 out of the 5 to be non major, we have to sum all the previous results, and divide them by the ways we can choose the 5 students (C1):

P = 2,002 + 7,007 + 7,644 + 3,185 + 490 / 20,349

P = 0.9989

5 0
4 years ago
What is 129 divided by 12
marissa [1.9K]
129 divided by 12 is 10.75. hope this helps.
8 0
3 years ago
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Find the following GCD and LCM: (3.a) GCD(343,550), LCM(343, 550). (3.b) GCD(89, 110), LCM(89, 110). (3.c) GCD(870, 222), LCM(87
Ierofanga [76]

Answer with Step-by-step explanation:

(3.a) GCD(343,550), LCM(343, 550).

343=7×7×7

550=5×5×2×11

GCD(343,550)=1

LCM(343,550)=7×7×7×5×5×2×11=188650

(3.b) GCD(89, 110), LCM(89, 110).

89=1×89

110=5×2×11

GCD(89, 110)=1

LCM(89, 110)=89×5×2×11=9790

(3.c) GCD(870, 222), LCM(870, 222).

870=2×3×5×29

222=2×3×37

GCD(870, 222)=2×3=6

LCM(870, 222)=2×3×5×29×37=32190

5 0
4 years ago
Thank u to whom ever deleted that QUSTION
Arisa [49]

Answer:

Ummm wasn't me but thank you for the points hope you have a great holiday and new year!

Step-by-step explanation:

3 0
3 years ago
The factorial 4! is equal to _______.<br><br> This is about permutations btw.
bekas [8.4K]

Answer:

24

Step-by-step explanation:

factorial(4) = 4*3*2*1

factorial(4) = 24

8 0
4 years ago
Read 2 more answers
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