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yulyashka [42]
2 years ago
15

Math help rn please

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

Answer:

7

Step-by-step explanation:

He's reading at a rate of 1/3 chapter per night and if there 2 1/3 chapters left to read we can divide the chapters left by his reading rate (2 1/3 ÷ 1/3 = 7 nights).

Another way to solve this problem is to make an equation to show the situation with x being the number of nights reading.

1/3x=2 1/3

<em>Solve for x by dividng 1/3 from each side of the equation.</em>

x=7

So he must read for 7 nights to finish the chapters that are left.

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( \frac{4}{7y^5} )^2
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Hal has 4/8 yard of string. How many 1/12 yard pieces of string can he cut from 3/4 yard?​
Mrac [35]

Answer:

9

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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

So then the best option is : c. A and C

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