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2 years ago

A school survey reported that

Mathematics

1 answer:

kozerog [31]2 years ago

6 0

The probability that the student has a part-time job, given that they have a cell phone is 5/8

<h3>What is probability</h3>

Probability is the likelihood or chance that an event will occur.

Given the following parameter:

Total student = 80%
Part-time jobs = 45%
Those with both a cell phone and a part-time job = 30%

Student will cellphone only = 80 - 30 = 50%

The probability that the student has a part-time job, given that they have a cell phone is 50/80 = 5/8

Learn more on probability here: brainly.com/question/25870256

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Help me with this algebra problem. Thank you :)

nata0808 [166]

To solve this problem you must apply the proccedure shown below:

1. You must make a system of equations, as below:

2. Let's call:

x: the number of first class tickets bought.
y: the number of coach tickets bought.

3. Then, you have:

x+y=7 (First equation)
970x+370y=3790 (Second equation)

x=7-y

4. By substitution, you have:

970x+370y=3790
970(7-y)+370=3790
y=5

x=7-y
x=7-5
x=2

Therefore, the answer is:

- Number of first class tickets bought=2

- Number of coach tickets bought=5

5 0

3 years ago

Is y = 2x - 3 a function

Lostsunrise [7]

Answer: Yes it is a function.

Step-by-step explanation: When trying to find out, you put it into a calculator and can tell by the line if it is a function.

7 0

2 years ago

At a large Midwestern university, a simple random sample of 100 entering freshmen in 1993 found that 20 of the sampled freshmen

guajiro [1.7K]

Answer:

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

p1 -> 1993

20 out of 100, so:

$p_1 = \frac{20}{100} = 0.2$

$s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04$

p2 -> 1997

10 out of 100, so:

$p_2 = \frac{10}{100} = 0.1$

$s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03$

Distribution of p1 – p2:

$p = p_1 - p_2 = 0.2 - 0.1 = 0.1$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.04^2 + 0.03^2} = 0.05$

Confidence interval:

$p \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower bound of the interval is:

$p - zs = 0.1 - 1.645*0.05 = 0.01775
$

The upper bound of the interval is:

$p + zs = 0.1 + 1.645*0.05 = 0.18225
$

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

6 0

3 years ago

Juliana wants to remodel her bathroom. Materials will cost $500 and the contractor charges $40 per hour. The equation y=40x+500

svet-max [94.6K]

what are you trying to find or how long did the worker work

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2 years ago

There were 78 girls trying out for dance. They were placed into groups containing d dancers. How many groups were there, in term

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2 years ago