Suppose you invest 16000 at 9% interest and that it is compounded daily. How much will you have in 8 years?

A square has a perimeter of 36x - 12. Which expression represents one side of the<br> square?

valina [46]

Answer:

9x - 3

Step-by-step explanation:

Hello!

A square has 4 sides.

We can factor out 4 from the expression to find the value of one side.

Factor:

36x - 12
4(9x) - 4(3)
4(9x - 3)

This means that each side of the square is 9x - 3 units long.

6 0

2 years ago

a hardware stores mixes paint in a ratio of three parts blue to five parts yellow to make two gallons of lime green. A ration of

Rufina [12.5K]

7B+5Y=487B+1Y=40Subtracting Eqn 2 from 1 we get 4Y=8so Y = $2 and X=$38/7

4 0

3 years ago

Using traditional methods, it takes 10.1 hours to receive a basic driving license. A new license training method using Computer

Law Incorporation [45]

Answer:

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis: $\mu =10.1$

Alternative hypothesis: $\mu \neq 10.1$

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$

$p_v =2*P(t_{(11)}>0.990)=0.343$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=10.5$ represent the sample mean

$s=1.4$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =10.1$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis: $\mu =10.1$

Alternative hypothesis: $\mu \neq 10.1$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=12-1=11$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(11)}>0.990)=0.343$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.

7 0

3 years ago

In Noels survey, 3 more people said they enjoyed volleyball than said they like soccer, and 5 more people enjoyed basketball tha

ExtremeBDS [4]

4 people liked soccer.

4 people liked tennis.

7 people like volleyball.

9 people like basketball.

3 0

3 years ago

The average Act score follows a normal distribution, with a mean of 21.1 and a standard deviation of 5.1. What is the probabilit

ohaa [14]

Answer:

0.43% probability that the mean IQ score of 50 randomly selected people will be more than 23

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 $s = \frac{\sigma}{\sqrt{n}}$ .