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salantis [7]
3 years ago
6

If the total number of people attending the game was 64,000, how many people were

Mathematics
1 answer:
guapka [62]3 years ago
6 0
Is this the whole question ?
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A certain university has 10 vehicles available for use by faculty and staff. four of these are vans and 6 are cars. on a particu
REY [17]
<span>There are 4 vans. So we have that probability that the first vehicle is a van p (e) = 4/10 = 0.4. P(e|f) = P( f and e) / p (f) p(e) = 0.4 and p(f) = 3/9 P (f and e) = 0.40 * 0.33 = 0.132 So p(e|f) = 0.4 * 0.33/ 0.33 = 0.4 P(f and e) p(f) * p(e) = 0.4 * 0.33 = 0.132</span>
6 0
3 years ago
Find the probability of exactly four successes in five trials of a binomial experiment in which the probability of success is 40
hodyreva [135]

Answer:

p(4 successes) ≈ 7.7%

Step-by-step explanation:

p(k successes in n trials) = C(n,k)·p^k·(1-p)^(n-k)

You have n=5, k=4, p=0.4, so the probability is ...

p(4 successes) = C(5,4)·(0.4)^4·(1 -0.4)^1 = 5·0.4^4·0.6 = 0.0768

p(4 successes) ≈ 7.7%

8 0
3 years ago
Read 2 more answers
HELP!!!
ella [17]

Answer:

120

Step-by-step explanation:

We are given that the function for the number of students enrolled in a new course is f(x) = 4^{x}-1.

It is asked to find the average increase in the number of students enrolled per hour between 2 to 4 hours.

We know that the average rate of change is given by,

A = \frac{f(x)-f(a)}{x-a},

where f(x)-f(a) is the change in the function as the input value (x-a) changes.

Now, the number of students enrolled at 4 = f(4) = f(x) = 4^{4}-1 = 255 and the number of students enrolled at 2 = f(2) = f(x) = 4^{2}-1 = 15

So, the average increase A=\frac{f(4)-f(2)}{4-2} = A=\frac{255-15}{4-2} = A=\frac{240}{2} = 120.

Hence, the average increase in the number of students enrolled is 120.

3 0
4 years ago
Someone just please help me no links no guessing seriously
Doss [256]

Answer:

no I can't u just ruined my mood

8 0
3 years ago
There are 6 green marbles, 11
algol13

Step-by-step explanation:

green: 6/4

red: 7/4

this is all i know

8 0
3 years ago
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