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saul85 [17]
3 years ago
5

Evaluate C(6, 6). A. Undefined B. 0 C. 1

Mathematics
2 answers:
irakobra [83]3 years ago
6 0
Evaluate C (6,6)
Answer: A. Undefined
3241004551 [841]3 years ago
5 0

I believe the answer is A. Undefined

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PLEASEEE SOMEONE HELP ME ON THIS!!!! It’s due in a couple hours
zimovet [89]

Answer:

x = 8.0 m

Step-by-step explanation:

With the angle presented, I'm assuming you know basic trigonometry. We know cosθ = adj/hyp from SOH CAH TOA. We have angle and hypotenuse, so plugging in those values we get cos(20°) = x / 8.5. Solve for x to get 8.5cos(20°) = 7.987387277 m = 8.0 m.

4 0
3 years ago
Read 2 more answers
Leo dug up 6 4/5 pounds of
Maurinko [17]
6 4/5 = 6.8
4 1/2 = 4.5

6.8-4.5= 2.3

Answer: 2 3/10 more pounds of treasure
7 0
3 years ago
Below is a probability distribution for the number of failures in an elementary statistics course. X 0 1 2 3 4 P(X=x) 0.41 0.18
faust18 [17]

Answer:

a) P(X=2)= 0.29

b) P(X<2)= 0.59

c) P(X≤2)= 0.88

d) P(X>2)= 0.12

e) P(X=1 or X=4)= 0.24

f) P(1≤X≤4)= 0.59

Step-by-step explanation:

a) P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4)= 1-0.41-0.18-0.06-0.06= 0.29

b) P(X<2)= P(X=0) + P(X=1)= 0.41 + 0.18 = 0.59

c) P(X≤2)= P(X=0) + P(X=1) + P(X=2)=0.41+0.18+0.29= 0.88

d) P(X>2)=P(X=3) + P(X=4)=0.06+0.06= 0.12

e) P(X=1 or X=4)=P(X=1 ∪ X=4) = P(X=1) + P(X=4)=0.18+0.06= 0.24

f) P(1≤X≤4)=P(X=1) + P(X=2) + P(X=3) + P(X=4)=0.18+0.29+0.06+0.06= 0.59

3 0
4 years ago
Read 2 more answers
Please help someone
ratelena [41]

Answer:

1. Y

2. N

3. N

4. N

Step-by-step explanation:

Let's use the second equation, since it seems to be easier to use.

To check if an ordered pair is a solution, plug it in to the equation.

1. -10+18=8 --> Y

2. 25-12=13 --> N

3. 0-9=-9 --> N

4. 35-27=8 --> N

Edit : The 4th equation doesn't work for the first equation, whereas the first one still does.

7 0
3 years ago
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Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose P(V) = 0.17
ArbitrLikvidat [17]

Answer: 0.82

Step-by-step explanation:

The probability of the computer not containing neither a virus nor a worm is expressed as P(V^{C}∩W^{C}) , where P(V^{C}) is the probability that the event V doesn't happen and P(W^{C}) is the probability that the event W doesn't happen.

P(V^{C})= 1-P(V) = 1-0.17 = 0.83

P(W^{C})=1-P(W) = 1-0.05 = 0.95

Since V^{C} and W^{C} aren't mutually exclusive events, then:

P(V^{C}∪W^{C}) = P(V^{C}) + P(W^{C}) - P(V^{C}∩W^{C})

Isolating the probability that interests us:

P(V^{C}∩W^{C})= P(V^{C}) + P(W^{C}) - P(V^{C}∪W^{C})

Where P(V^{C}∪W^{C}) = 1 - 0.04 = 0.96

Finally:

P(V^{C}∩W^{C}) = 0.83 + 0.95 - 0.96 = 0.82

5 0
3 years ago
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