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Deffense [45]
2 years ago
10

UHHH HELP OMG WHAT IS THIS I WASNT PAYING ATTENTION SORRY 26 POINTS DO NOT GIVE ME AN ANSWER IF U DONT KNOW HELP i know its easy

but im lost

Mathematics
1 answer:
Karolina [17]2 years ago
3 0

Answer:

m=2/3, b=12

Step-by-step explanation:

When you see an equation in this format, it's called slope-intercept form (y=mx+b, with m being slope and b being y-intercept). So in this case, by looking at our formula, we can tell right away that our slope, or m, is 2/3 and our y-intercept, or b, is 12. HTH :)

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Evaluate 0.00048×0.81×10×(10)-7 ÷0.027×0.04x(10)6​
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Answer:

Step-by-step explanation:

0.00048*.81=0.0003888

0.0003888*10=0.003888

0.003888*(10)=0.3888

0.3888-7=-6.6112

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The Magazine Mass Marketing Company has received 1515 entries in its latest sweepstakes. They know that the probability of recei
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Answer:

P(x < 5)=P(X=0)+P(X=1)+P(X=2)+P(x=3)+P(X=4)

P(X=0)=(15C0)(0.5)^0 (1-0.5)^{15-0}=0.0000305

P(X=1)=(15C1)(0.5)^1 (1-0.5)^{15-1}=0.000458

P(X=2)=(15C2)(0.5)^2 (1-0.5)^{15-2}=0.0032

P(X=3)=(15C3)(0.5)^3 (1-0.5)^{15-3}=0.0139

P(X=4)=(15C4)(0.5)^4 (1-0.5)^{15-4}=0.0417

And adding we got:

P(x < 5)=0.0592

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=15, p=0.5)

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

And we want to find this probability:

P(x < 5)=P(X=0)+P(X=1)+P(X=2)+P(x=3)+P(X=4)

P(X=0)=(15C0)(0.5)^0 (1-0.5)^{15-0}=0.0000305

P(X=1)=(15C1)(0.5)^1 (1-0.5)^{15-1}=0.000458

P(X=2)=(15C2)(0.5)^2 (1-0.5)^{15-2}=0.0032

P(X=3)=(15C3)(0.5)^3 (1-0.5)^{15-3}=0.0139

P(X=4)=(15C4)(0.5)^4 (1-0.5)^{15-4}=0.0417

And adding we got:

P(x < 5)=0.0592

7 0
3 years ago
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