Write an equation for a line perpendicular to y=2x-4 and passes through the point (-2,0)

Lena [83]

678.987Answer:

Step-by-step explanation:

5 0

2 years ago

7. The value of a computer has a decay factor of 0.85 per year. After 2 years, the computer is worth $1445. What was the origina

Readme [11.4K]

Answer:

$64222.22

Step-by-step explanation:

If we call the original value as x, we can write:

1445 = x * (0.15)² (Note: We write 1 - 0.85 = 0.15 and not 0.85 because 0.85 is the decay factor)

1445 = 0.0225x

x ≈ 64222.22

7 0

3 years ago

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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

What is the check digit(d) for the UPC bar code 7-25272-73070-d A)8 B)3 C)4 D)2

vovangra [49]

B I hope this help I think it’s right

4 0

2 years ago

There are less than 200 apples in a box. it is known that 2, 3, 4, 5, or 6 kids can share these apples evenly. how many apples c

MArishka [77]

Hi!

200 ÷ 2 = 50

200 ÷ 3 = 66.6 repeat.

200 ÷ 4 = 50

200 ÷ 5 = 40

200 ÷ 6 = 33.3 repeat.

So, there can be either 50 or 40 apples in that box.

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2 years ago

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