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3 years ago

13

Deivn brought his snail collection to school he has 10 snail how could he put them into 2 tank so two class could see them write

the equation for all the possible way

1 answer:

ioda3 years ago

4 0

He could put 5 in one and another 5 in the other. So just divide so he could put 2 snails in 5 different tanks and so on

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The Information Technology Department at a large university wishes to estimate the proportion of students living in the dormitor

nadya68 [22]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16$

And rounded up we have that n=385

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can use as an estimator for p $\hat p =0.5$ . And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16$

And rounded up we have that n=385

3 0

3 years ago

A bag of 8 rolls for $1.89 or a bag of 18<br> rolls for $3.79

Delvig [45]

Answer:

8 rolls for $1.89

Step-by-step explanation:

when you divide to see how much each one costs the 8 rolls is better

4 0

3 years ago

Read 2 more answers

2 1⁄2 ÷ 1 3⁄6 please help me with the RIGHT answer.

ycow [4]

Answer:

$1\frac{2}{3}$

Step-by-step explanation:

We want to simplify:

$2 \frac{1}{2} \div 1 \frac{3}{6}$

This is the same as:

$2 \frac{1}{2} \div 1 \frac{1}{2}$

Now let us convert the mixed numbers to improper fractions.

$\frac{5}{2} \div \frac{3}{2}$

We multiply by the reciprocal of the second fraction to get:

$\frac{5}{2} \times \frac{2}{3}$

Cancel out the common factors to obtain:

$\frac{5}{3} = 1 \frac{2}{3}$

5 0

3 years ago

Which of the following is not one of the three ways to express ratio A 3/4 B 3:4 C 3 of 4 D 3 to 4

serious [3.7K]

3 of 4 is not the correct way of writing a ratio.

7 0

3 years ago

Read 2 more answers

The employees of a firm that manufactures insulation are being tested for indications of asbestos in their lungs. The firm is re

balandron [24]

Answer:

0.013 = 1.3% probability that fifteen employees must be tested in order to find three positives.

Step-by-step explanation:

For each employee, there are only two possible outcomes. Either they test positive, or they do not. The probability of an employee testing positive is independent of any other employee, which means that the binomial probability distribution is used to solve this question.

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.

40% of the employees have positive indications of asbestos in their lungs

This means that $p = 0.4$

Find the probability that fifteen employees must be tested in order to find three positives.

2 during the first 14(given by P(X = 2) when n = 14).

The 15th is positive, with 0.4 probability. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{14,2}.(0.4)^{2}.(0.6)^{12} = 0.0317$

0.0317*0.4 = 0.013.

0.013 = 1.3% probability that fifteen employees must be tested in order to find three positives.

7 0

2 years ago