How to get the answer to this question

Math but tell me don’t make me click on som link

Alisiya [41]

Answer:

theres no question lol

Step-by-step explanation:

4 0

3 years ago

Need help on this question thank you

Ivahew [28]

Answer:

B.

Step-by-step explanation:

8 0

4 years ago

Researchers sample 9000 households concerning TV shows they watch. Based on the sample 14% reported watching 60 Minutes. What i

svetlana [45]

Answer:

What is centripetal acceleration?

Can an object accelerate if it's moving with constant speed? Yup! Many people find this counter-intuitive at first because they forget that changes in the direction of motion of an object—even if the object is maintaining a constant speed—still count as acceleration.

Acceleration is a change in velocity, either in its magnitude—i.e., speed—or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant. You experience this acceleration yourself when you turn a corner in your car—if you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we'll examine the direction and magnitude of that acceleration.

The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration

8 0

4 years ago

When the store had sold 2/9 of the boots that were on display, they brought out another 34 pairs from the stock room. If that ga

nikdorinn [45]

Let us assume number of boots pairs were on display originally = x.

We are given 2/9 of the total boots pairs we sold.

So, 2/9 of x boots pairs were sold, that is

Number of boots pairs were sold = 2/3 x.

Number of boots pairs brought out from store = 34.

Remaining pairs of boots = 174 pairs.

So, we can setup an equation,

Total number of boots pairs - First number of boots pairs sold + number of boots pairs brought out from store = Remaining pairs.

x - 2/9 x +34 = 174.

Subtracting 34 on both sides, we get

x - 2/9 x = 34-34 = 174-34

x- 2/9 x = 140 .

Simplifying left fracions by taking common denominator,

9/9 x- 2/9 x = 140

(9x-2x)/9 = 140.

7/9 x = 140.

Multiplying both sides by 9.

9* 7/9 x = 9*140.

7x = 1260.

Dividing both sides by 7.

7x/7 = 1260/7

x= 180.

Therefore, 180 pairs were on displayed originally.

5 0

4 years ago

Find the P-value for the indicated hypothesis test. An airline claims that the no-show rate for passengers booked on its flights

trapecia [35]

Answer:

$p_v =P(z$

B. 0.2061

Step-by-step explanation:

1) Data given and notation

n=380 represent the random sample taken

X=19 represent the no-shows in the sample

$\hat p=\frac{19}{380}=0.05$ estimated proportion of no-shows in the sample

$p_o=0.06$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the no-show rate for passengers booked on its flights is less than 6%:

Null hypothesis: $p\geq 0.06$

Alternative hypothesis: $p < 0.06$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.05-0.06}{\sqrt{\frac{0.06(1-0.06)}{380}}}=-0.8208$

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 next step would be calculate the p value for this test.

Since is a one left tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very high value and using the significance level for example $\alpha=0.05$ 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 no-show rate for passengers booked on the company flights is nor significantly less than 0.06 or 6%.

4 0

3 years ago

How to get the answer to this question​

How to get the answer to this question