Please help me I can't figure this out.

John runs a computer software store. He counted 122 people who walked by his store in a day, 60 of whom came into the stores. Of

ozzi

Answer:

unlikely

Step-by-step explanation:

60/122

60 ÷122=0.49

0.49 × 100 =49

49% chance

answer is unlikely

5 0

3 years ago

Can someone please help as soon as possible!!

sergiy2304 [10]

Answer:

1) triangles are similar

Step-by-step explanation:

The height from vertex X of isosceles ∆WXY is 4 units. The width WY is also 4 units. In isosceles ∆UVW, the height from vertex V is 6 units, and the width UW is also 6 units.

The height ratios are ...

∆WXY/∆UVW = 4/6

The width ratios are ...

∆WXY/∆UVW = 4/6

The measures of ∆WXY are proportional to those of ∆UVW, so the triangles are similar.

_____

Strictly speaking, you cannot go by triangle height and width alone. That is why we made not of the fact that the triangles are <em>isosceles</em>. When base and height of an isosceles triangle are proportional, the Pythagorean theorem guarantees that side lengths are proportional. Trigonometry can also be invoked to support the claim that angles are congruent.

4 0

3 years ago

5. Using the information from a previous problem and given the tax table, find

Leno4ka [110]

Answer:

Step-by-step explanation:

.............................. where is the table

3 0

2 years ago

Erythromycin is a drug that has been proposed to possibly lower the risk of premature delivery. A related area of interest is it

katrin2010 [14]

Answer:

$z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01$

$p_v =P(z>1.01)=0.156$

So the p value obtained was a very low value and using the significance level asumed $\alpha=0.05$ 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 women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

Step-by-step explanation:

Data given and notation

n=195 represent the random sample taken

X=65 represent the women who complain of nausea between the 24th and 28th week of pregnancy

$\hat p=\frac{65}{195}=0.333$ estimated proportion of women who complain of nausea between the 24th and 28th week of pregnancy

$p_o=0.3$ is the value that we want to test

$\alpha$ represent the significance level

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 higher than 0.3.:

Null hypothesis: $p\leq 0.3$

Alternative hypothesis: $p > 0.3$

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:

$z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01$

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 assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>1.01)=0.156$

So the p value obtained was a very low value and using the significance level asumed $\alpha=0.05$ 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 women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

6 0

3 years ago

At noon, a private plane left Austin, Texas for Los Angeles, California. The distance between Austin and LA is 2100 km. The plan

Oxana [17]

Answer:

Infinity

Step-by-step explanation:

1+1=5

2+2=22 =

infinity

6 0

2 years ago