What is 0.15% of 250? in the simplest form

matrenka [14]

Huh, this is presented pretty interesting to me.

To start with, if you actually had no idea what the answer to this question was, you wouldn't pick B or C because x has to be close to 91°. In this particular quadrilateral, it's especially important that we focus on the fact, "x - 15." This means that x has to be higher initially to be subracted to equal 91, or the opposite angle.

In order to subtract to 91, the answer would have to be 106, or D. This is a hunch, but I have a suspicion that angle B and D are going to be the same, being that their on opposite sides of the figure.

Hope this helps! If you have further questions, let me know.

4 0

3 years ago

A manufacturer of a new medication on the market for Alzheimer's disease makes a claim that the medication is effective in 65% o

NeX [460]

Answer:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

Step-by-step explanation:

Data given and notation

n=180 represent the random sample taken

X=115 represent the adults with the medication was effective

$\hat p=\frac{115}{180}=0.639$ estimated proportion of adults with the medication was effective

$p_o=0.65$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is less than 0.65.:

Null hypothesis: $p \geq 0.65$

Alternative hypothesis: $p < 0.65$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.639 -0.65}{\sqrt{\frac{0.65(1-0.65)}{180}}}=-0.309$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults with the medication was effective is not significantly less than 0.65

3 0

3 years ago

A(n) _________ is a fraction that describes a comparison of two numbers or quantities.

jolli1 [7]

An ratio is a fraction that describes a comparison of two numbers or quanitites

3 0

3 years ago

Frn3n ionfrnffn kfrk jfrnfn

kvasek [131]

I tried to help but couldn’t :)

5 0

3 years ago

Need a step by step explanation pls help me

AlladinOne [14]

Answer:

This is my guess...

"Even today, the need for work is a common reason people move to urban areas."

Step-by-step explanation:

The article talks about technology and how it lead to bigger cities in the past. Because we are reading about prior years, there is no need to talk about the present day.

7 0

3 years ago