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snow_tiger [21]
3 years ago
6

To win at LOTTO in one state, one must correctly select 6 numbers from a collection of 62 numbers (1 through 62). The order in w

hich the selection is made does not matter. How many different selections are possible? ...?
Mathematics
1 answer:
marissa [1.9K]3 years ago
5 0
We can use the concept of combinations for this particular problem since it was clearly stated that the order in which the selection is made does not matter. By using a calculator, we can obtain the value of 62C6 of 61474519. So, there are 61474519 possible different selections. 
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Solve for m:<br> (-11/6) + m = -2/9<br> M=
ss7ja [257]

Step-by-step explanation:

(-11/6) + m = -2/9

m = -2/9 + 11/6

the last common multiple of 9 and 6 is 18.

so, we are bringing both fractions to .../18.

m = -2×2/18 + 11×3/18 = -4/18 + 33/18 = 29/18

8 0
2 years ago
Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive
PtichkaEL [24]
Looks like the PMF is supposed to be

\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65

Next,

\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25

\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5

\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2
\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5
\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}

If Y=X^2+1, then X^2=Y-1\implies X=\sqrt{Y-1}, where we take the positive root because we know X can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that Y can take on the values 1^2+1=2, 2^2+1=5, and 5^2+1=26. At these values of Y, we would have the same probability as we did for the respective value of X. That is,

\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}

Part (5) is incomplete, so I'll stop here.
8 0
3 years ago
Cookies are sold singly or in packages of 3 or 9. With this​ packaging, how many ways can you buy 18 cookies?
8_murik_8 [283]
3 : 3 times 6 is 18, 2 times 9 is 18 and 3 times 3 is 9 plus 9 is 18
4 0
3 years ago
Estimate the square root of 50 to the hundredths place
Anarel [89]
We have the following expression:
 root (50)
 Rewriting we have:
 root (50) = root (2 * 25)
 root (2 * 25) = 5 * root (2)
 5 * root (2) = 7.071067812
 Round to the hundredths place:
 root (50) = 7.07
 Answer:
 
the square root of 50 to the hundredths place is:
 
7.07
5 0
3 years ago
Read 2 more answers
Suppose a parabola has vertex (–8, –7) and also passes through the point (–7, –4). Write the equation of the parabola in vertex
Nutka1998 [239]
That’s the answer and also the steps to getting that answer

6 0
3 years ago
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