Bonnie earns $455 a week, and she works 5 days a week. what is her daily wage?

Of the 800 participants in a marathon 220 are running to raise money for a cause. How many participants out of 100 are running f

nadezda [96]

Answer:

27.5 /100 are running for a cause

Step-by-step explanation:

We know that 220/800 are running for a cause

Divide the top and bottom by 8 to get the fraction out of 100

220 /8 =27.5

800/8 = 100

27.5 /100 are running for a cause

7 0

3 years ago

Help me please!! i am already failing the test

Ivan

Answer:

2x²+2x

Step-by-step explanation:

you have f(x)= x+1 and g(x)= 2x

you must multiply the 2 polynomials

(1x·2x) +( 1·2x)

1·2=2 x^1·x^1=x²

2x²

1·2x=2x

4 0

3 years ago

Plz help me!! Polynomial Long Division (Level 2)

____ [38]

Answer: 3x^2 - 3x - 1

Explanation:

I got a picture to show

7 0

3 years ago

How to solve 4y=x 3x- y=70 using substitution

creativ13 [48]

X = 4y
3x - y = 70

substitute the x in the second equation with the other half of the first equation.

3(4y) - y = 70
12y - y = 70
11y = 70
y = 6.36 repeating, or 6 4/11

now use this to solve first equation
x = 4 (6 4/11)
x = 25.45 repeating

done

7 0

3 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