Ask question

Ask question

All categories

3 years ago

7

An online news source reports that the proportion of smartphone owners who use a certain operating system is 0.216. Two simulati

ons will be conducted to create a sampling distribution of the sample proportion from a population with proportion 0.216. Simulation A will consist of 1,000 trials with a sample size of 50. Simulation B will consist of 500 trials with a sample size of 100.

1 answer:

kaheart [24]3 years ago

4 0

Answer:

The center/ mean will almost be equal, and the variability of simulation B will be higher than the variability of simulation A.

Step-by-step explanation:

Solution

Normally, a distribution sample is mostly affected by sample size.

As a rule, sampling error decreases by half by increasing the sample size four times.

In this case, B sample is 2 times higher the A sample size.

Now, the Mean sampling error is affected and is not higher for A.

But it's sample is huge for this, Thus, they are almost equal

Variability of simulation decreases with increase in number of trials. A has less variability.

With increase number of trials, variability of simulation decreases, so A has less variability.

You might be interested in

Identify a temperature colder than -7.2f. Write an inequality that relates the temperatures. Describe their positions on a horiz

madreJ [45]

Colder then -7.2 F would be -8.5 F

-7.2 > -8.5

anything to the left of -7.2 on the number line will be less then(oe in this case, colder then) -7.2....so -8.5 falls to the left of -7.2

8 0

3 years ago

A credit card contains 16 digits between 0 and 9. However, only 100 million numbers are valid. If a number is entered randomly,

marusya05 [52]

Answer:

$P(valid) = 1 * 10^{-8}$

Step-by-step explanation:

Let us first find the number of possible credit card numbers that can be formed with the numbers 0 - 9.

We want to form a 16 digit number out of 10 digits, so, we have:

$N = 10^{16}$

There are $10^{16}$ numbers that can be formed.

There are only 100 million valid numbers ( $100 * 10^6 = 10^8$ ). Let the number of valid numbers be n.

Therefore, the probability of picking a number that is valid is:

P(valid) = n / N

$P(valid) = 10^{8} / 10^{16}\\\\P(valid) = 1 * 10^{-8}$

8 0

3 years ago

Estimate the fractional part of the shaded portions below. Explain how you decided

zloy xaker [14]

Answer:

do you have a picture?

8 0

4 years ago

Read 2 more answers

I need a word problem or this asap. THANK YOU!!!!!!!!!<br> x + y ≤ 40<br> 4x + 5y ≥80

CaHeK987 [17]

Carson needed pencil and erasers, but the total amount can not be over 40. Pencils were $4 and erasers were 5$. How much pencils and erasers can he buy without spender over $80

7 0

3 years ago

Which shows one way to determine the factors of 4x3 + x2 – 8x – 2 by grouping?

Korvikt [17]

Remember that to factoring by grouping, we should split the expression into tow pairs that have something in common. The idea is that after factoring each pair, we get a binomial common factor. Is worth pointing out that you can pair factors in more than one way.

Let's check how to determine the factors of our expression by grouping:

Step 1. Look for common multiples between the terms.

Notice that 4 and 8 are multiples of 2, 8 is a multiple of 4, and all of them are multiples of 1, so we are going to try more than one pairing combination.

Step 2. Look for common factors between the variables.

Usually, we want to pair variables with higher exponents so when we factor each pair, we end up with another common factor in binomial form.

Step 3. Factor each pair; if you end with another binomial common factor, you nailed it. If not, try again.

Les't apply our steps to our expression $4x^3+x^2-8x-2$

Applying step 1, we are pairing $4x^3$ and $-8x$ . Remember to pair the remaining factors as well.

$(4x^3-8x)+(x^2-2)$

The common factor in the first pair is $4x$ , so lets factor that out

$4x(x^2-2)+(x^2-2)$

Look! we have the binomial common factor $(x^2-2)$ . Unfortunately, $4x(x^2-2)+(x^2-2)$ is not in our given choices, so let's try again:

This time we are going to group the higher variables (step 2) $4x^3$ and $x^2$

$4x^3+x^2-8x-2$

$(4x^3+x^2)+(-8x-2)$

The common factor in the first pair is $x^2$ , and the common factor in the second one is -2, so let's factor out both of them:

$(4x^3+x^2)+(-8x-2)$

$x^2(4x+1)-2(4x+1)$

This time, we got option A.

We can conclude that the correct answer is: A) x2(4x + 1) – 2(4x + 1)

3 0

3 years ago