How to write 84 divided into sevenths

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.70 m/s2. at the same inst

77julia77 [94]

You know that t=0 and x=0, the car starts moving with a constant acceleration of 2.70m/s²

At the same time, a truck moving speed o 8.50 m/s passes the car

D=v.t (for the truck)

And d= v.t+1/2.a.t²(for the car)

v.t=1/2.a.t²

t=2.v/a

t=2.8.50 m/s/ 2.70 m/s²

17m/2.70=6.29

This mean that the distance is

D=v.t=8.50m/s.6.29= 53.465 m/s

The speed of the car is

Vcar=a.t

2.7m/s2×× 53.65

=144.34

4 0

3 years ago

Look at Arnold's attempt to solve the equation for b:

Tasya [4]

His mistake was in multiplying the 3 with the 12. He should have divided it from the 12 because that would have gotten rid of the 3 with the variable. It would also lead to the correct answer, which is b=4

4 0

4 years ago

You estimate that there are 66 cars in a parking lot. The actual number of cars is 75.

Lubov Fominskaja [6]

IDK but go to my pro file to help youuuuuuuuuuuuuuuuuuuuuuu

7 0

3 years ago

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

What's the sum off 663 and 23

vlada-n [284]

Answer: 686

Step-by-step explanation: 663 + 23 = 686

8 0

2 years ago