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garik1379 [7]
2 years ago
14

Chelsea is graphing the function f(x) = 20(One-fourth)x. She begins by plotting the initial value. Which graph represents her fi

rst step?
Mathematics
2 answers:
natta225 [31]2 years ago
8 0

Answer:

C

Step-by-step explanation:

Alenkinab [10]2 years ago
4 0

Answer:

c

Step-by-step explanation:

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20 cups yogurt and 30 cups blueberry
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Circle groups to show 3x(2x3)
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2x3=6x3=18 u always do PEMDAS,
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Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y > a) = qa .
lakkis [162]

Answer:

a) For this case we can find the cumulative distribution function first:

F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k

So then by the complement rule we have this:

P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a

b) P(Y>a)= q^a

P(Y>b) = q^b

So then we have this using independence:

P(Y> a+b) = q^{a+b}

We want to find the following probability:

P(Y> a+b |Y>a)

Using the definition of conditional probability we got:

P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)

And we see that if a = 2 and b=5 we have:

P(Y> 2+5 | Y>2) = P(Y>5)

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

P(X=x)=(1-p)^{x-1} p

If we define the random of variable Y we know that:

Y\sim Geo (1-p)

Part a

For this case we can find the cumulative distribution function first:

F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k

So then by the complement rule we have this:

P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a

Part b

For this case we can use the result from part a to conclude that:

P(Y>a)= q^a

P(Y>b) = q^b

So then we have this assuming independence:

P(Y> a+b) = q^{a+b}

We want to find the following probability:

P(Y> a+b |Y>a)

Using the definition of conditional probability we got:

P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)

And we see that if a = 2 and b=5 we have:

P(Y> 2+5 | Y>2) = P(Y>5)

Part c

For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

8 0
2 years ago
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. x+y = 3, x = 4
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The intersection between the curves are
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The volume is
V = 108π/5 or 67.86
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Which number in the monomial 125x18y3z25 needs to be changed to make it a perfect cube?
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