Circular flower garden 18 feet in diameter. calculate area in square feet.

It is claimed that automobiles are driven on average more than 20,000 kilometers per year. To test this claim, 100 randomly sele

kow [346]

Answer:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.

Step-by-step explanation:

1) Data given and notation

$\bar X=23500$ represent the sample mean

$s=3900$ represent the sample standard deviation

$n=100$ sample size

$\mu_o =2000$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mean is higher than 20000, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2000$

Alternative hypothesis: $\mu > 2000$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=100-1=99$

Since is a one-side upper test the p value would be:

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.

8 0

3 years ago

The range of a relation is....

maxonik [38]

its simple X is domain Y is range so the answer is

A

8 0

3 years ago

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In an engineering test, a rocket sled is propelled into a target. The sled’s distance d in meters from the target is given by th

creativ13 [48]

You're given d = 10
10 = -1.5t<span>2 = 110
</span> $t^{2}$ = 220/3
since t $\geq$ 0
t = $\sqrt 220/3$
I apologize if this is wrong!

6 0

3 years ago

a meteorologist measures the angle of elevation of a weather balloon with the ground as 41. A radio signal from the balloon indi

Aneli [31]

Here you would use right triangle trig (SOH CAH TOA)
So first draw a right triangle. Imagine youre standing at the angle opposite the right angle which is the one on the ground.This angle is the 41°. Now Imagine the balloon is as the angle above the right triangle. Well since the balloon is 1503 m from his location this would be the hypotenuse. SInce we are trying to find the height (x) we would use sine since
sine = opposite/hypotenuse . Now lets solve make sure calculator is in degree mode:

sin41 = x/1503 multiply both sides by 1503 to cancel it out
1503sin41 = x plug into calculator
x = 986.057 ft
The balloon is 986.057 feet above the ground.

6 0

3 years ago

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Plot the line for the equation on thr graph. y-2=-2(x+4)

julia-pushkina [17]

Answer:

see attached

Step-by-step explanation:

y-2=-2(x+4)

y -2 = x(-2) + 4(-2)

y -2 = -2x - 8

y = -2x - 8 + 2

y = -2x - 6

When x = 0, y = -6. Hence (0,-6) is a point on the line

when y = 0, x = -3. Hence (-3,0) is a point on the line.

Plot these 2 points and connect with a line (see attached)

6 0

3 years ago