Jonah read 5 1/2 chapters in his book in 90 minutes how long did it take him to read one chapter

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Answer 1

Answer:

Answer:

around 16 minutes. you partition an hour and a half (all out) by what number of sections he read (5.5

Step-by-step explanation:

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NBC News reported on May 2, 2013, that 1 in 20 children in the United States have a food allergy of some sort. Consider selecting a random sample of 15 children and let X be the number in the sample who have a food allergy. Then X ~ Bin(15, 0.05). (Round your probabilities to three decimal places.) (a) Determine both P(X ≤ 3) and P(X < 3). P(X ≤ 3) = P(X < 3) = (b) Determine P(X ≥ 4). P(X ≥ 4) = (c) Determine P(1 ≤ X ≤ 3). P(1 ≤ X ≤ 3) = (d) What are E(X) and σX? (Round your answers to two decimal places.) E(X) = σX = (e) In a sample of 90 children, what is the probability that none has a food allergy?

Answers

Answer:

a) P(X ≤ 3) = 0.9946

P(X < 3) = 0.9639

b) P(X ≥ 4) = 0.0054

c) P(1 ≤ X ≤ 3) = 0.5313

d) E(X) = 0.75

σX = 0.84

e) P(X=0) = 0.0099

Step-by-step explanation:

We have x: number in the sample who have a food allergy. As the sample is of n=15 and p=0.05, we have:

$X \sim Bin(15, 0.05)$

a) We have to determine P(X ≤ 3) and P(X < 3)

We can calculate P(X ≤ 3) as the sum of P(0), P(1), P(2) and P(3).

$P(x\leq 3)=\sum_(k=0)^3P(k)\n\n\nP(x=0) = \binom{15}{0} p^(0)q^(15)=1*1*0.4633=0.4633\n\nP(x=1) = \binom{15}{1} p^(1)q^(14)=15*0.05*0.4877=0.3658\n\nP(x=2) = \binom{15}{2} p^(2)q^(13)=105*0.0025*0.5133=0.1348\n\nP(x=3) = \binom{15}{3} p^(3)q^(12)=455*0.0001*0.5404=0.0307\n\n\nP(x\leq 3)=0.4633+0.3658+0.1348+0.0307=0.9946$

P(x<3) can be calculated from the previos result as:

$P(x<3)=P(X\leq3)-P(3)=0.9946-0.0307=0.9639$

b) We can calculate P(X ≥ 4) as:

$P(X\geq4)=1-P(X<4)=1-P(X\leq3)=1-0.9946=0.0054$

c) We can calculate P(1 ≤ X ≤ 3) as:

$P(1 \leq X \leq 3)=P(1)+P(2)+P(3)=0.3658+0.1348+0.0307=0.5313$

d) The expected value of a binomial variable is the product of the sample size n and the probability of success p:

$E(X)=np=15*0.05=0.75$

The standard deviation is calculates as:

$\sigma_x=√(np(1-p))=√(15*0.05*0.95)=√(0.7125) =0.84$

e) In this case, the sample size is n=90.

We can calculate the probability that none has a food allergy as:

$P(x=0) = \binom{90}{0} p^(0)q^(90)=0.95^(90)=0.0099$

6.solve an inequality that represents the description and then solve Toni can carry up to 18 lb in her backpack.
Her lunch weighs 1 lb, her gym clothes weigh
2 lb, and her books (b) weigh 3 lb each. How
many books can she carry in her backpack?

Answers

Answer:

nothing

Step-by-step explanation:

One normally distributed with mean value 20 in. and standard deviation .5 in. The length of the second piece is a normal rv with mean and standard deviation 15 in. and .4 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation .1 in. Assuming that the lengths and amount of overlap are independent of each other, what is the probability that the total length after insertion is between 34.5 and 35 in.

Answers

Answer:

The probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.

Step-by-step explanation:

Let the random variable X represent the length of the first piece, Y represent the length of the second piece and Z represents the overlap.

It is provided that:

$X\sim N(20,\ 0.50^(2))\nY\sim N(15,\ 0.40^(2))\nZ\sim N(1,\ 0.10^(2))$

It is provided that the lengths and amount of overlap are independent of each other.

Compute the mean and standard deviation of total length as follows:

$E(T)=E(X+Y-Z)\n=E(X)+E(Y)-E(Z)\n=20+15-1\n=34$

$SD(T)=√(V(X+Y-Z))\n=√(V(X)+V(Y)+V(Z))\n=\sqrt{0.50^(2)+0.40^(2)+0.10^(2)}\n=0.6480741\n\approx 0.65$

Since X, Y and Z all follow a Normal distribution, the random variable T, representing the total length will also follow a normal distribution.

$T\sim N(34, 0.65^(2))$

Compute the probability that the the total length after insertion is between 34.5 and 35 inches as follows:

$P(34.5<T<35)=P((34.5-34)/(0.65)<(T-\mu_(T))/(\sigma_(T))<(35-34)/(0.65))\n\n=P(0.77<Z<1.54)\n\n=P(Z<1.54)-P(Z<0.77)\n\n=0.93822-0.77935\n\n=0.15887\n\n\approx 0.1589$

*Use a z-table.

Thus, the probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.

On Wednesday, a local hamburger shop sold a combined total of 360 hamburgers and cheeseburgers. The number of cheeseburgers sold was two times the number of hamburgers sold. How many hamburgers were sold on Wednesday?

Answers

Answer:

120

Step-by-step explanation:

360 combined, 2/3 is cheese, so 1/3 is hamburger, 1/3 of 360 is 120.

Please help the first people were wrong!!!

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Answer:

The surface area is about 267.04 cm^2; the surface area would be divided by 4 if the slant height and diameter were to be cut in half

A researcher wants to determine if birthweights of children born to U.S. mothers is affected in any way when large megadoses of caffeine are consumed routinely by the mother during pregnancy. It is known that the birthweights are normally distributed with mean 7.5 pounds and standard deviation 1 pound. When drawing a sample of size 40 from such a population, and computing the mean birthweight in the sample, use the Central Limit Theorem to find the 2.5-th percentile of the distribution of sample means.

Answers

Answer:

7.19

Step-by-step explanation:

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