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Debora [2.8K]
4 years ago
13

Rewrite 7.13 as a mixed number in lowest terms.

Mathematics
1 answer:
Anit [1.1K]4 years ago
3 0
The lowest 7.13 can go as a mixed number is 7 13/100.
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____________________
JulijaS [17]
3/4J - 1 = 16 <== ur equation
3 0
4 years ago
If two fair dice are rolled, what is the probability that the total showing is either even or less than nine?
Gemiola [76]

Answer:

1/2 and 13/18 respectively

Step-by-step explanation:

even-number results:

2, 4 6 ,8, 10 and 12

the combinations of each result are respectively

1+3+5+5+3+1 = 18

The total number of combinations on a roll pair is 36

then the probability of obtaining an even result is:

p = 18/36 = 1/2 or 0.5

so that the result is less than nine, we eliminate the combinations of results 9, 10, 11 and 12:

4 + 3 + 2 + 1 = 10

36 -10 = 26

the probability of obtaining a result less than nine is:

p = 26/36 = 13/18 or 0.72

Hope this helps

5 0
2 years ago
A building has n floors numbered 1,2,...,n, plus a ground floor g. at the ground floor, m people get on the elevator together, a
fomenos
Let X_i be the random variable indicating whether the elevator does not stop at floor i, with

X_i=\begin{cases}1&\text{if the elevator does not stop at floor }i\\0&\text{otherwise}\end{cases}

Let Y be the random variable representing the number of floors at which the elevator does not stop. Then

Y=X_1+X_2+\cdots+X_{n-1}+X_n

We want to find \mathrm{Var}(Y). By definition,

\mathrm{Var}(Y)=\mathbb E[(Y-\mathbb E[Y])^2]=\mathbb E[Y^2]-\mathbb E[Y]^2

As stated in the question, there is a \dfrac1n probability that any one person will get off at floor n (here, n refers to any of the n total floors, not just the top floor). Then the probability that a person will not get off at floor n is 1-\dfrac1n. There are m people in the elevator, so the probability that not a single one gets off at floor n is \left(1-\dfrac1n\right)^m.

So,

\mathbb P(X_i=x)\begin{cases}\left(1-\dfrac1n\right)^m&\text{for }x=1\\\\1-\left(1-\dfrac1n\right)^m&\text{for }x=0\end{cases}

which means

\mathbb E[Y]=\mathbb E\left[\displaystyle\sum_{i=1}^nX_i\right]=\displaystyle\sum_{i=1}^n\mathbb E[X_i]=\sum_{i=1}^n\left(1\cdot\left(1-\dfrac1n\right)^m+0\cdot\left(1-\left(1-\dfrac1n\right)^m\right)
\implies\mathbb E[Y]=n\left(1-\dfrac1n\right)^m

and

\mathbb E[Y^2]=\mathbb E\left[\left(\displaystyle\sum_{i=1}^n{X_i}\right)^2\right]=\mathbb E\left[\displaystyle\sum_{i=1}^n{X_i}^2+2\sum_{1\le i

Computing \mathbb E[{X_i}^2] is trivial since it's the same as \mathbb E[X_i]. (Do you see why?)

Next, we want to find the expected value of the following random variable, when i\neq j:

X_iX_j=\begin{cases}1&\text{if }X_i=1\text{ and }X_j=1\\0&\text{otherwise}\end{cases}

If X_iX_j=0, we don't care; when we compute \mathbb E[X_iX_j], the contributing terms will vanish. We only want to see what happens when both floors are not visited.

\mathbb P(X_iX_j=1)=\left(1-\dfrac2n\right)^m
\implies\mathbb E[X_iX_j]=\left(1-\dfrac2n\right)^m
\implies2\displaystyle\sum_{1\le i

where we multiply by n(n-1) because that's how many ways there are of choosing indices i,j for X_iX_j such that 1\le i.

So,

\mathrm{Var}[Y]=n\left(1-\dfrac1n\right)^m+2n(n-1)\left(1-\dfrac2n\right)^m-n^2\left(1-\dfrac1n\right)^{2m}
4 0
3 years ago
Pythagorean theorem 3d
gayaneshka [121]

Answer:

ihhrhi2rhhb4rhbur4hbur4hb43hubr4u3br

Step-by-step explanation:

3 0
3 years ago
Eight less than the quotient of a number and 3 is 18. Which equation models this sentence?
labwork [276]
B- n/ 3 - 8 = 18....................
8 0
3 years ago
Read 2 more answers
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