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marin [14]
3 years ago
13

What is the theoretical probability of rolling a sum of 8 on one roll of two standard number cubes

Mathematics
1 answer:
Vesnalui [34]3 years ago
3 0

I just took the test, the answer is 5/36.

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5 x (10^2) is equal to: 450 4,500 45,000 450,000
vazorg [7]
5 x (10^2) =
5 x 100 = 500

I guess you mean:
4,5 x (10^2) =
4,5 x 100 =
450

Which acctually is a valid answer.
4 0
3 years ago
A club with 31 members is to select five officers (president, vice president, secretary, treasurer, and historian). In how many
Ksivusya [100]
If you take 31*5 that equals = 155 ways! <span />
5 0
3 years ago
Suppose that in one metropolitan area, 25% of all home- owners are insured against earthquake damage. Four home-owners are to be
emmasim [6.3K]

Answer:

Step-by-step explanation:

Given that,

25% home owners are insecure of earthquake problem

If we select 4 home owners at random

let X denote the number among the four who have earthquake insurance

Let find the probability distribution of X

Let S- denotes a home owner who has insurance

Let F denotes a home owner who does not have insurance.

Then, P(S) =25% = 0.25

P(F) = 1 — P(S) = 1 —0.25

P(F) = 0.75

The possible outcomes of X is

X(0) = If no person has insurance

X(1) = If only 1 person has insurance

X(2) = If only 2 persons has insurance

X(3) = If only 3 persons has insurance

X(4) = If only 4 persons has insurance.

The cardinality of the sample space is n(C) = 2ⁿ = 2⁴ => 16

So, the sample space is given as

{FFFF, FFFS, FFSF, FFSS, FSFF, FSFS, FSSF, FSSS, SFFF, SFFS, SFSF, SFSS, SSFF, SSFS, SSSF, SSSS}

For X=0, the possible is {FFFF} i.e. no insurance, the one without insurance.

P(X) = 0.25×0.25×0.25×0.25

P(X) = 0.25⁴

P(X) = 0.00390625

For X=1, the possible outcome are

FFFS, FFSF, FSFF, SFFF

P(X=1) = 0.25³•0.75 + 0.25³•0.75 + 0.25³•0.75 + 0.25³•0.75

P(X=1) = 0.046875

For X=2 the possible outcomes are

FFSS, FSFS, FSSF, SFFS, SFSF, SSFF,

P(X=2)=0.25²•0.75²+0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²

P(X=2) = 0.2109375

For X=3 the possible outcomes are

FSSS, SFSS, SSFS, SSSF

P(X=3) = 0.25•0.75³+0.25•0.75³+ 0.25•0.75³+0.25•0.75³

P(X=3) = 0.421875

X=4 the possible outcomes are

SSSS

P(X=4) = 0.75×0.75×0.75×0.75

P(X=4) = 0.31640625.

Or

Using normal distribution

P(X=k) = ⁿCk • 0.25^k • 0.75^(4-k)

So,

P(X=0) = 4C0 • 0.25^0 • 0.75^4

P(X=0) = 0.31640625

P(X=1) = 4C1 • 0.25^1 • 0.75^3

P(X=1) = 0.421875

P(X=2) = 4C2 • 0.25^2 • 0.75^2

P(X=2) = 0.2109375

P(X=3) = 4C3 • 0.25^3 • 0.75^1

P(X=3) = 0.046875

P(X=4) = 4C4 • 0.25^4 • 0.75^0

P(X=4) = 0.00390625.

6 0
3 years ago
Determine the height of the can of beans(cylinder) given the volume of the
jeka94

Answer:

4 in(a)

Step-by-step explanation:

height-?

volume-314

radius-5

pie-22÷7 or 3.14

volume= πrsquared h

314=3.14×5×5×h

314= 78.5h

314÷78.5

= 4 in

8 0
2 years ago
The school band sold 3,305 raffle tickets to raise money for a trip to Disney World. Which equation shows the relationship betwe
Iteru [2.4K]
I believe it would be D.
4 0
3 years ago
Read 2 more answers
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