Whats 1 plus 1 pls help im lonley

The temperature in Alaska can reach -30 degrees Fahrenheit during the winter months it is colder in Antarctica than in the Alask

Fofino [41]

Answer:

B. -40°F

Step-by-step explanation:

its- clearly lower than -30

4 0

2 years ago

Read 2 more answers

Which of the following represents x =1/3y written in general form?

Serhud [2]

If you would like to write x = 1/3 * y in general form, you can do this using the following steps:

The general form of the equation is: ax + by + c = 0.

x = 1/3 * y

x - 1/3 * y = 0

The correct result would be x - 1/3 * y = 0.

4 0

3 years ago

Sam average 13 points per game in her first three basketball games. She averaged 14.5 points in her first four games. How many p

kvasek [131]

Answer:

19

Step-by-step explanation:

13×3= 39

14.5×4= 58

58-39= 19

5 0

3 years ago

Y=3x-5 y=-6x+4 system of equations

slega [8]

X=1, y=-2--------------------------

6 0

3 years ago

In a recent poll, 778 adults were asked to identify their favorite seat when they fly, and 492 of them chose a window seat. Use

seropon [69]

Answer:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

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

$p_v =P(Z>7.36)=9.19x10^{-14}$

Since the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .

Step-by-step explanation:

1) Data given and notation

n=778 represent the random sample taken

X=492 represent the people that chose a window seat.

$\hat p=\frac{492}{778}=0.632$ estimated proportion of people that chose a window seat.

$p_o=0.5$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the majority of adults prefer window seats when they fly:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.632 -0.5}{\sqrt{\frac{0.5(1-0.5)}{778}}}=7.36$

4) 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 right tailed test the p value would be:

$p_v =P(Z>7.36)=9.19x10^{-14}$

5) Conclusion

Since the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .

6 0

3 years ago