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erastova [34]
3 years ago
9

How would you put that in an equation?

Mathematics
1 answer:
Jet001 [13]3 years ago
6 0

Answer:

I don't see anything here attach first

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Let N be a geometric random variable with parameter p. Given N, generate N many i.i.d. random numbers U1, U2, . . . , UN uniform
erma4kov [3.2K]

Answer:

The correct answer is x^np (1-p)^n-1

Step-by-step explanation:

See the picture attached

5 0
3 years ago
10) Thirty-seven percent of the American population has blood type O+. What is the probability that at least four of the next fi
lisabon 2012 [21]

Answer:

0.06597

Step-by-step explanation:

Given that thirty-seven percent of the American population has blood type O+

Five Americans are tested for blood group.

Assuming these five Americans are not related, we can say that each person is independent of the other to have O+ blood group.

Also probability of any one having this blood group = p = 0.37

So X no of Americans out of five who were having this blood group is binomial with p =0.37 and n =5

Required probability

=The probability that at least four of the next five Americans tested will have blood type O+

= P(X\geq 4)\\= P(X=4)+P(x=5)\\= 5C4 (0.37)^4 (1-0.37) + 5C5 (0.37)^5\\= 0.06597

7 0
3 years ago
Read 2 more answers
If 6 cans of soup cost $1.50, how much will 9 cans cost?
Step2247 [10]

If 6 cans of soup cost $1.50

9 cans cost .......................x?

x = 9 * 1.50 / 6

x = 2.25

Answer

9 cans cost $2.25

3 0
3 years ago
What is the mean for the following set of data?<br> 2,5, 7, 8, 11, 11, 12<br> 10<br> 11<br> 8
Yuri [45]
Add up all the numbers & divide by the number of numbers. That should give you the mean.
4 0
3 years ago
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An economy pack of highlighters contains 12 yellow, 6 blue, 4 green, and 3 orange highlighters. An experiment consists of random
Serhud [2]

Answer:

The probability of choosing a blue or yellow highlighter when one yellow highlighter is picked is 17/25.

Step-by-step explanation:

Here, according to the question:

Total number of yellow highlighters  = 12

Total number of blue  highlighters  = 6

Total number of green highlighters  = 4

Total number of orange  highlighters  = 3

So, the total number of highlighters  =  12 + 6 + 4 + 3  = 25

Let E1 : Event of selecting a yellow highlighter.

P(E1)  = \frac{\textrm{Total number of yellow highlighters}}{\textrm{Total Highlighters}}  = \frac{12}{25}

Let E2 : Event of selecting blue or yellow highlighter ONCE yellow highlighter is selected.

So, the total yellow highlighters left after selecting one = 12 -1  = 11

Also, the number of blue highlighter  =  6

So, the TOTAL FAVORABLE options to E2 = 6+ 11 = 17

P(E2)  =

\frac{\textrm{Total number of yellow  + Blue   highlighters}}{\textrm{Total Highlighters}} = \frac{17}{25}

Hence, the probability that a blue or yellow highlighter is selected given that a yellow highlighter is selected is 17/25.

7 0
3 years ago
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