Factorization of 4x+7-x

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

PLEASE HELP! Image below

maria [59]

Answer:

1.6

Step-by-step explanation:

So, first, write down the number of jokers in each deck of cards.

That would be: 0, 1, 2, 2, 3

Then, add up all those numbers.

0+1+2+2+3=8

Now, divide that sum by the number of decks or cards shown in the data set, which would be 5.

8/5=1.6

So, the mean of the data in the dot plot shown is 1.6.

:L)

4 0

3 years ago

HELP NEEDED!

Ilia_Sergeevich [38]

Answer:

x = 12.6

Step-by-step explanation:

Sum of measure of all the arcs of a circle = 360°

m(arc CD) + m(arc CED) = 360°

Since, m(arc CED) = (20x)°

And m(arc CD) = (8x + 6)°

By substituting these measures in teh expression,

(20x)° + (8x + 6)° = 360°

(20x + 8x) + 6 = 360

28x = 360 - 6

28x = 354

x = $\frac{354}{28}$

x = 12.64

x = 12.6

4 0

3 years ago

What is the solution?

Ipatiy [6.2K]

Answer:

3.) {2x+2y=62; x+4y=70

4 0

3 years ago

Someone please help! ATTACHMENT (just the circled one)

Elza [17]

4 (3x+9)-3(5y+7) =11

distribute

12x +36 -15y -21 =11

12x -15y +15 =11

subtract 15 from each side

12x -15y =-4

subtract 12x from each side

-15y= - 12x-4

divide by -15

y = -12/-15 x -4/-15

y = 4/5 x + 4/15

slope = 4/5

y intercept = 4/15

6 0

3 years ago