Here Is My Last Question I Took A Snip Of 30 Points

Find acceleration of a cheetah running at 0.5 and sect 21 and second three seconds describe the acceleration of the animal.

xenn [34]

Answer:

Initial velocity (u)=o m/s

Final velocity (v)=27m/s

Time taken to attain velocity is(t)=5m/s

As we know v=u+at

Substitute above values in the given equation

27=0+a(5)

27/5=a

a=5.4 m/s^2

Step-by-step explanation:

4 0

2 years ago

Pls help answer before 10:00

antiseptic1488 [7]

Answer:

1. 14

2. 2.75

3. 7.25

4. 6.5

Step-by-step explanation:

1. You just make a dot plot and count the amount of dots there are.

2. You go to your dot plot and see the smallest amount and you get your answer.

3. Same thing as the smallest amount, except you just have to find the greatest amount.

4. You just find which number appears the most on the dot plot, and bam you got your answer.

6 0

3 years ago

PLEASE HELP URGENT DUE IN 2 HOURS

Kaylis [27]

Answer:

look it up on slader.com thats the sams problem that i have had before and it worked on slader.

8 0

2 years ago

A quadratic function y=f(x) is plotted on a graph and the vertex of the resulting parabola is (−5,−3). What is the vertex of the

Licemer1 [7]

Answer:

(-5 , 6)

Step-by-step explanation:

f(x) = (x - (-5))² - 3 ... vertex (h , k) : (-5 , -3) y = (x-h)² + k

g(x) = -f(x) + 3 = -((x+5)² - 3) + 3 = -(x + 5)² + 6

vertex of g(x) is : (-5 , 6)

4 0

2 years ago

The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) At α = .05 i

fenix001 [56]

Answer:

Null hypothesis: $\mu \geq 250$

Alternative hypothesis: $\mu < 250$

$p_v =P(t_{15}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=510$ represent the mean for the sample

$s=50$ represent the standard deviation for the sample

$n=16$ sample size

$\mu_o =530$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:

Null hypothesis: $\mu \geq 530$

Alternative hypothesis: $\mu < 530$

We don't know the population deviation and the sample size is lees than 30, 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:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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".

3) Calculate the statistic

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

$t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60$

4) Calculate the critical value

We need to begin calculating the degrees of freedom

$df=n-1=16-1=15$

The critical value for this case would be :

$P(t_{15}$

The value of a that satisfy this on the normal standard distribution is a=-1.75 and would be the critical value on this case zc=-1.75

5) Calculate the P-value

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

$p_v =P(t_{15}$

6) Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

8 0

2 years ago