Please help me with this question!

Attila the Hun was a ruler of the Hunnic Empire. The Romans feared him greatly until his death since he made quite a habit of in

Bas_tet [7]

The inequality would be x > 453 because any year after he died would be more than the year he died.
Also, if it helps, I think of inequalities as an alligator's mouth, trying to eat the bigger quantity.
Hope this helps!

8 0

3 years ago

What is the remainder when 849,400 is divided by 216

S_A_V [24]

The anserws is 88 so your welcome

5 0

3 years ago

Read 2 more answers

The secret number has a 7 in the ones place. The tens number is an odd number

Lana71 [14]

This really is left up to you as long as you have an odd number in the tens place and a 7 in the ones place. How big (or small) the number is is up to you. As long as it is at least 17, it's pretty much up to you.

7 0

3 years ago

Which is equivalent to (4xy-3x)^2, and what type of special product is it?

mars1129 [50]

It's not the difference of squares, rather it is the square of a difference. That leaves a perfect square trinomial, which narrows your selection to two choices. An expression with 2 terms is not a trinomial, so that further narrows your selection. The appropriate choice is

... (4xy -3z)² = 16x²y² -24xyz +9z², a perfect square trinomial

_____

The expression you have in your problem statement has no z term, so none of the choices is applicable to that one.

8 0

3 years ago

Read 2 more answers

A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second po

igor_vitrenko [27]

Answer:

a) For this case the value of the significanceis $\alpha=0.05$ and $\alpha/2 =0.025$ , we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

$z_{\alpha/2} =1.96$

If the calculated statistic $|z_{calc}| >1.96$ we can reject the null hypothesis at 5% of significance

b) Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8$

c) $z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708$

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Step-by-step explanation:

Data given and notation

$X_{1}=170$ represent the number of people with the characteristic 1

$X_{2}=110$ represent the number of people with the characteristic 2

$n_{1}=200$ sample 1 selected

$n_{2}=150$ sample 2 selected

$p_{1}=\frac{170}{200}=0.85$ represent the proportion estimated for the sample 1

$p_{2}=\frac{110}{150}=0.733$ represent the proportion estimated for the sample 2

$\hat p$ represent the pooled estimate of p

z would represent the statistic (variable of interest)

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

$\alpha=0.05$ significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

a.State the decision rule.

For this case the value of the significanceis $\alpha=0.05$ and $\alpha/2 =0.025$ , we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

$z_{\alpha/2} =1.96$

If the calculated statistic $|z_{calc}| >1.96$ we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8$

c. Compute the value of the test statistic.

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708$

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

5 0

3 years ago