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3 years ago

8

Madison is reading a book that has 232 pages in it. She read 42 pages over the weekend. Then she read 30 more pages Monday night

. How many pages does Madison have left to read?

2 answers:

denpristay [2]3 years ago

8 0

Answer:

160

Step-by-step explanation:

You add the amount of pages that she has read in all which is 42+32=72. Then you subtract the sum by the amount fo pages in the book: 232-72 which equals 160

Lelechka [254]3 years ago

3 0

Answer:

160

Step-by-step explanation:

She read 42 pages over the weekend and 30 more pages Monday night. Add those together. The result is 72.

Do 232 - 72. You get 160.

In an equation, it looks like this: 232 - (42 + 30) = 160

Hope this helped! :)

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Consider the probability that greater than 26 out of 124 software users will call technical support. Assume the probability that

Mekhanik [1.2K]

Answer:

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

100% probability that greater than 26 out of 124 software users will call technical support.

Step-by-step explanation:

Test if the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

It is needed that:

$np \geq 10$ and $n(1-p) \geq 10$

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Out of 124 software users

This means that $n = 124$

Assume the probability that a given software user will call technical support is 97%.

This means that $p = 0.97$

Conditions:

$np = 124*0.97 = 120.28 \geq 10$

$n(1-p) = 124*0.03 = 3.72 < 10$

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

Consider the probability that greater than 26 out of 124 software users will call technical support.

The lowest possible probability of those is 27, so, if it is 0, since it is considerably below the mean, 100% probability of being greater. We have that:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 27) = C_{124,27}.(0.97)^{27}.(0.03)^{97} = 0$

1 - 0 = 1

100% probability that greater than 26 out of 124 software users will call technical support.

3 0

3 years ago

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Serga [27]

It’s either 64 or 192 if that helps because it say she buys a three week ad would u times 64 by three I’m sorry I don’t have a final answer

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3 years ago

Eric and Mark were at a ball game and visited the concession stand. They each got a hamburger, and Mark got lemonade and some co

EastWind [94]

So here is the answer. Given that both Eric and Mark got a hamburger which is $4 each. Mark got a lemonade and some cotton candy which costs $5.
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3 years ago

The average and standard deviation of a measurement is 15.0 ± 5.0. if 100 students measured the value, how many would you expect

AURORKA [14]

Solution: We are given:

$\mu=15, \sigma=5$

Using the empirical rule, we have:

$\mu \pm \sigma$ covers 68% of data.

Also the percentage of values below mean = Percentage of values above mean = 50%

Now, let's find the z score for x=20

$z=\frac{20-15}{5}=1$

Therefore, the percentage of values greater than 1 standard deviation above mean $50\% - \frac{68\%}{2} =50\%-34\%=16%$

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3 years ago

A wildlife shelter is home to birds, mammals, and reptiles. If cat chow is sold in 30 lb bags, what is the least number of bags

kotykmax [81]

Answer:

26 bags

Step-by-step explanation:

A bag of cat chows = 30 lbs

Amount of cat chows needed in a week = 15 lbs = ½ bag of cat chow

There are at least 52 weeks in a year.

The least number of cat chows bags needed at the shelter in 1 year = 52 weeks × ½ bag of chow

= 52 × ½

= 52/2

= 26 bags of cat chow

At least, 26 bags would be needed in 1 year (52 weeks) at this wild life shelter if 15 lbs is consumed weekly.

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2 years ago

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