Use the numbers to write related addition and subtraction sentences 6 7 8 9

ALL of the Students at Beach Middle School voted for a trip to the zoo or the aquarium.

QveST [7]

Answer:

160

Step-by-step explanation:

If there were only 2 options, all the people who didn't pick one chose the other one.

So if 25% of the people voted for the aquarium the remaing 75% voted for the zoo.

Since three times as many people voted for the zoo, we can see that 120/3=40, so 40 people chose the aquarium.

To find the total number of students, we add 120 and 40 to get <u>160</u>

3 0

1 year ago

2) Working alone, Rob can dig a 10 ft by 10 ft hole in ten hours. Shayna can dig the same hole in seven hours. Find how long it

Free_Kalibri [48]

rob can dig 1/10ft in one hour

Shayna can dig 1/7ft in one our

together they can dig

1/10+1/7

(7+10)/70 in one hour

17/70

in for hours they can dig

17/70×4

68/70

the left work is 2/70

in 60 minutes they can dig 17/70

1/70 work can be done in 60 minutes/17

2/70 work can be done in 60/17×2 minutes

7.088 minutes

for the entire work they take 4hours and 7 minutes

8 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

The graph shows the girth of the trunk of a fir tree after different numbers of years: What does the y-intercept of the graph re

Talja [164]

The y-intercept represents the initial girth of the tree. Assuming that the X-axis represents time in this scenario (as it usually does) the y-axis, therefore, represents the girth of the tree. The y-intercept is located where X = 0. So if no time has passed then this must be the first measurement of the girth of the tree.

5 0

2 years ago

I dont understand this question

mamaluj [8]

I have no clue what grade are you in

8 0

3 years ago