Need help with 2 math questions pls help me for 20 points.

Find the distance between the pair of points.<br> (-5,2) (2,-2)

Nonamiya [84]

(7,4) The x distance is 7 units and the y distance is 4 units. Hope it helps.

3 0

3 years ago

A cylinder shaped juice pitcher has a diameter of 12 cm and a height of 25 cm. What volume of juice does the pitcher contain whe

Andreas93 [3]

Answer:

706.50 cubic cm.

Step-by-step explanation:

Given that a cylinder shaped juice pitcher has a diameter of 12 cm and a height of 25 cm.

We are to find out the t volume of juice does the pitcher contain when it is 25% full. 3.14 is to be used for pi.

Volume of the cylinder = $\pi r^2 h\\ \\=3.14 *6^2*25\\=2826$

When it is 25% full the volume would be 1/4 of the full volume

= $\frac{2826}{4} =706.50$ cubic cm

So volume when 25% full is

706.50 cubic cm.

4 0

4 years ago

A store pays $306.56 for a refrigerator. The store marks up the price by 29%. What is the amount of the mark-up?

umka21 [38]

The price of the mark up would be 395.4624

I got this by doing:
306.56 x 29/100 = 88.9024
88.9024 + 306.56 = 395.4624

7 0

3 years ago

Read 2 more answers

X² - 6x + 4 = 0<br><br> solve and show work. ...?

ycow [4]

The answer is 3 + √5 and 3 - √5.

This is quadratic equation: x² - 6x + 4 = 0
The general quadratic equation is ax² + bx + c = 0
Our equation can be rewritten: 1x² + (-6)x + 4 = 0
So, in our equation we have: a = 1, b = -6, c = 4

Now, x can be calculated using the formula: $x_{1,2}= \frac{-b+/- \sqrt{ b^{2}-4ac } }{2a}$
$x_{1,2} =\frac{-(-6)+/- \sqrt{ (-6)^{2}-4*1*4 } }{2*1} = \frac{6+/- \sqrt{36-16} }{2} = \frac{6+/- \sqrt{20} }{2} =\frac{6+/- \sqrt{4*5} }{2}= \\ 
=\frac{6+/- \sqrt{4}* \sqrt{5} }{2} =\frac{2*3+/- 2\sqrt{5} }{2}= \frac{2(3+/- \sqrt{5}) }{2} =3+/- \sqrt{5}$

From here:
$x_{1} =3+ \sqrt{5} \\ 
x_{2} =3- \sqrt{5}$

6 0

4 years ago

Last year, a comprehensive report stated that 28% of businesses in the northeast of Ohio were considered highly profitable. This

Alik [6]

Answer:

$z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575$

$p_v =2*P(z>1.575)=0.115$

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ 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 businesses were highly profitable is not significantly different from 0.28 or 28%.

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

X=19 represent the businesses were highly profitable

$\hat p=\frac{19}{50}=0.38$ estimated proportion of businesses were highly profitable

$p_o=0.28$ 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 the true proportion of businesses were highly profitable is different from 0.28 or no, the system of hypothesis is.:

Null hypothesis: $p=0.28$

Alternative hypothesis: $p \neq 0.28$

When we conduct a proportion test we need to use the z statistic, 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 required we can replace in formula (1) like this:

$z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575$

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 $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>1.575)=0.115$

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ 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 businesses were highly profitable is not significantly different from 0.28 or 28%.

4 0

4 years ago