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dimulka [17.4K]
3 years ago
9

HELP PLEASE AND THANK YOU!!!! ASAP VERY IMPORTANT!!!!!!!!!!!!

Mathematics
1 answer:
gayaneshka [121]3 years ago
7 0

Answer:

1 xx

Step-by-step explanation:

each box got to to worth 1 point representing the x variable so i added 1 plus 1xx and minus 2 to get 1xx

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Zepler [3.9K]

Answer:

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option c is the correct

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A landscaper is planting a row of 10 shrubs along the walkway shown below. There must be one shrub at the very beginning and one
rewona [7]

Answer:

9d = 36

Step-by-step explanation:

See attachment for complete question

Length = 36ft

n = 10

From the question, we understand that one shrub must be at both ends.

Taking the first as a point of reference, we're left with:

(n - 1) * distance = 36

Substitute 10 for n and d for distance

(10 - 1) * d = 36

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Hence:

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4 0
4 years ago
in a certain population, 11% of people are left-handed. Suppose that you plan to randomly select 100 people and ask each person
Assoli18 [71]

Answer:

c. A and C

Step-by-step explanation:

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".

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

X \sim Binom(n=100, p=0.11)

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!}

We need to check the conditions in order to use the normal approximation.

np=100*0.11=11 > 10 \geq 10

n(1-p)=100*(1-0.11)=99 \geq 10

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

E(X)=np=100*0.11=11

\sigma=\sqrt{np(1-p)}=\sqrt{100*0.11(1-0.11)}=3.129

Part A

We want this probability:

P(X \geq 12) = 1-P(X

The z score is defined as

Z=\frac{x-\mu}{\sigma}.

P(X \geq 12) = 1-P(X

Part B

P(X>12) = 1-P(X\leq 12) = 1-P(Z< \frac{12-11}{3.129})=1-0.625=0.375[/tex]

Part C

P(10\leq X \leq 14) = P(X

The z score is defined as

Z=\frac{x-\mu}{\sigma}.

P(10 \leq X \leq 14) =P(Z< \frac{14-11}{3.129}) -P(Z< \frac{10-11}{3.129})=P(Z

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