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scoray [572]
3 years ago
14

Hehe wanna gf lol.................

Mathematics
2 answers:
wlad13 [49]3 years ago
5 0

Answer:

Ermmm.............

Step-by-step explanation:

Aloiza [94]3 years ago
3 0

Answer:

Imagine

Step-by-step explanation:

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Eli, a pretty smart ape in the Salt Lake City zoo, has correctly picked the winner of the last 7 Super Bowls. Read about Eli's p
Anna [14]

Answer:

There is a 0.78% probability that Eli chooses 7 winners in a row.

Step-by-step explanation:

There were 7 Super Bowls.

In each Super Bowl, there were two teams.

So, the probability that Eli chooses 7 winners in a row is

P = \frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{128} = 0.0078

There is a 0.78% probability that Eli chooses 7 winners in a row.

4 0
3 years ago
(0) 14 -7 0<br> | () 14 - 7 0
lyudmila [28]

Answer:

What is the question.

5 0
4 years ago
A square park has a diagonal walkway from one corner to another. If the walkway is 120 meters long, what is the approximate leng
Vlad1618 [11]

Answer:

  85 m

Step-by-step explanation:

The diagonal of a square is √2 times the length of the side. The park will have a side length of 120/√2 m ≈ 84.85 m, about 85 meters.

_____

The relations are ...

  diagonal = (√2)×(side length)

  side length = diagonal/√2 . . . . . . . . . divide the above equation by √2

5 0
4 years ago
Pleaseee help I’m kinda confused!!!! Write equivalent expressions by combining like terms.
densk [106]
22x
It’s the same as 12 plus 10
8 0
3 years ago
4. At a local university 54.3% of incoming first-year students have computers. If 3 students are selected at random, and the fol
valentinak56 [21]

Answer:

a) 0.0954

b) 0.9045

c) 0.1601

Step-by-step explanation:

We are given the following information:

We treat first-year students having computers as a success.

P(first-year students have computers) = 54.3% = 0.543

Then the number of first year students follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 3

a) P(None have computers)

P(x =0) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3\\\\= 0.0954

b) P(At least one has a computer)

P(x \geq 1) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3+\binom{3}{1}(0.543)^1(1-0.543)^2+\binom{3}{3}(0.543)^3(1-0.543)^0\\\\=0.3402+0.4042+0.1601= 0.9045

c) P(All have computers)

P(x =3) \\\\= \binom{3}{3}(0.543)^3(1-0.543)^0\\\\= 0.1601

3 0
3 years ago
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