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Gwar [14]
3 years ago
5

Given that, in a certain school, the following are true, what is the probability that a student is taking both math and computer

science?
1. The probability that a student is taking math is 23%.
2. The probability that a student is taking computer science is 45%.
3. The probability that a student is taking math or computer science is 58%.

A) 8%
B) 11%
C) 68%
D) 81%
Mathematics
1 answer:
Delicious77 [7]3 years ago
8 0

The probability that a student is taking both math and computer science is option B) 11%.

<u>Step-by-step explanation:</u>

Given that,

The probability that a student is taking math alone = 23%.

The probability that a student is taking computer science alone = 45%.

The probability that a student is taking math or computer science = 58%.

<u>To find the probability that a student is taking both math and computer science :</u>

We use the formula P(A∪B) = P(A) + P(B) - P(A∩B)

Here,

P(math ∪ computer) = P(math) + P(computer) + P(math or computer)

⇒ 23% + 45% - 58%

⇒ 10%

Therefore, 10% is the probability that a student is taking both math and computer science.

We can choose option B) 11% as the correct answer since it is nearest to the 10%.

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artcher [175]

Answer:

A. Simple Random Sampling

B. Cluster Sampling

C. Convenience Sampling

D. Systematic Sampling

E. Stratified Sampling

Step-by-step explanation:

Simple Random Sampling is the sampling when samples are chosen randomly, where each unit has an equal chance of being selected in a sample.

If the population is divided into a different group called cluster and clusters are selected as a sample then it is Cluster Sampling.

In Convenience sampling, observers collect the sample as his\her convenience.

In Systematic Sampling sample is chosen by some criteria like he\she is taken every 10th unit as a sample from the population.

In Stratified Sampling population is divided into several groups such that within the group it is homogeneous and between the group it is heterogeneous. And now a selection of each stratum and unit has an equal chance of selection.

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3 years ago
Find the surface area of the composite solid. round the answer to the nearest hundredth.
lubasha [3.4K]
The picture in the attached figure

[surface area of the composite solid]=1*(4*6)+2*(4*4)+2*(4*6)+2*(4*√13/2)+2*(6*2√2/2)
[surface area of the composite solid]=24+32+48+4√13+12√2
[<span>surface area of the composite solid]=135.39 yd</span>²

the answer is
135.39 yd²

5 0
4 years ago
I need Help plz...!!!!
kirill [66]

Answer:

B.21.3 is the correct answer

Step-by-step explanation:

1/2*x+1/4*x=16

x/2+x/4=16

2x+x/4=16

3x/4=16

3x=16*4

3x=64

x=21.3

Hope it is helpful for you

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6 0
3 years ago
Read 2 more answers
3x - 2(2x - 5) = 2(x + 3) - 8
nataly862011 [7]

Answer:

x = 4

Step-by-step explanation:

Hello!

We can solve for x by expanding the parentheses and isolating x.

<h3>Solve for x</h3>
  • 3x - 2(2x - 5) = 2(x + 3)-8
  • 3x - 4x + 10 = 2x + 6 - 8
  • -x + 10 = 2x - 2
  • 10 = 3x - 2
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The value of x is 4.

3 0
2 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3
In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

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

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that \mu = 8.3, \sigma = 3.3.

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

s = \frac{3.3}{\sqrt{37}} = 0.5425

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

This probability is the pvalue of Z when X = 8.65. So

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

Z = \frac{8.65 - 8.3}{0.5425}

Z = 0.65

Z = 0.65 has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is subtracted by the pvalue of Z when X = 7.43

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

Z = \frac{7.43 - 8.3}{0.5425}

Z = -1.60

Z = -1.60 has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is the pvalue of Z when X = 8.65 subtracted by the pvalue of Z when X = 7.43.

So:

From a), we have that for X = 8.65, we have Z = 0.65, that has a pvalue of 0.7422.

From b), we have that for X = 7.43, we have Z = -1.60, that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0
3 years ago
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