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Anestetic [448]
2 years ago
15

An arts academy requires there to be 5 teachers for every 95 students and 4 tutors for every 40 students how many students does

the academy have per teacher? Per tutor
Mathematics
1 answer:
Whitepunk [10]2 years ago
5 0

Answer:

19 students per teacher and 10 students per tutor

Step-by-step explanation:

To find the amount of students per teacher and per tutor you have to divide

95 students /5 teachers

19 students per teacher

40 students /4 tutors

10 studetns pet tutor

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Make it 0 on one side <br><br> x2 - 2x + = 4(x2-2x +1)-3
Shkiper50 [21]

Answer:

-3x² +6x -1 = 0

Step-by-step explanation:

x2 - 2x + = 4(x2-2x +1)-3

x² - 2x = 4x²-8x +4 - 3

x²-4x²-2x+8x = 4-3

-3x² +6x = 1

-3x² +6x -1 = 0

4 0
3 years ago
Use the graph to find the cost of 6 show tickets.
Zinaida [17]
The lines y = 2x - 1 and y = -2x + 3 intersect at exactly one point which means this system has exactly one solution. so the system is consistent and independent 
3 0
3 years ago
the temperature started at 89°f. if the temperature went up 6°f and then down 12°f, how many degrees above or below the high is
Sonja [21]
The answer is 6 degrees below the high. 
89+6=95
95-12=83
4 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
This a comprehension questins​
nikdorinn [45]

where is the question?????????

3 0
3 years ago
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