Answer? please help :))

Nady [450]

Answer:

1. The y-intercept 8 means you start off with $8. The slope 1/4 means how much money you have to pay for the amount of tickets purchased for carnival games.

2. The y-intercept 2 means the number of chapters you started reading and the slope 5/4 means the number of chapters you read per day.

3. The y-intercept 3 means the inital (start) cost of a muffin delivery and the slope 2 means the cost of each muffin delivery for the number of muffins you ordered.

Number 3 on the slope part might be the other way around if I'm wrong :)

7 0

3 years ago

Estimate the unit rate to the nearest hundredth $2.35

Scilla [17]

It would be $2.40. 4 or less just ignore 5 or more add 1 more

8 0

3 years ago

Help me please and thx

Olenka [21]

Y = kx, so y/x is constant

you want y such that
y/12 = -2/-8

3 0

2 years ago

(10 points)Assume IQs of adults in a certain country are normally distributed with mean 100 and SD 15. Suppose a president, vice

vesna_86 [32]

Answer:

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Step-by-step explanation:

To solve this question, we need to use the binomial and the normal probability distributions.

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.

Normal Probability Distribution:

Problems of normal 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.

Probability the president will have an IQ of at least 107.5

IQs of adults in a certain country are normally distributed with mean 100 and SD 15, which means that $\mu = 100, \sigma = 15$

This probability is 1 subtracted by the p-value of Z when X = 107.5. So

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

$Z = \frac{107.5 - 100}{15}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 probability that the president will have an IQ of at least 107.5.

Probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

First, we find the probability of a single person having an IQ of at least 130, which is 1 subtracted by the p-value of Z when X = 130. So

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

$Z = \frac{130 - 100}{15}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772.

1 - 0.9772 = 0.0228.

Now, we find the probability of at least one person, from a set of 2, having an IQ of at least 130, which is found using the binomial distribution, with p = 0.0228 and n = 2, and we want:

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{2,0}.(0.9772)^{2}.(0.0228)^{0} = 0.9549$

$P(X \geq 1) = 1 - P(X = 0) = 0.0451$

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

What is the probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130?

0.3085 probability that the president will have an IQ of at least 107.5.

0.0451 probability that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

Independent events, so we multiply the probabilities.

0.3082*0.0451 = 0.0139

0.0139 = 1.39% probability that the president will have an IQ of at least 107.5 and that at least one of the other two leaders (vice president and/or secretary of state) will have an IQ of at least 130.

8 0

2 years ago

Marvin was buying soil for his flower garden. What would be the most reasonable amount of soil he would need to buy for a flower

muminat

Answer:

60 liters

Step-by-step explanation:

5 0

3 years ago

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