0.25% of 42 is what number

Hunter-Best [27]

Answer:

I believe its D , to increase the acceleration of a object you have to increase the force preferably ,the mass kind of slows it down because of the weight .

But d is the most legit answer so :p

hoped I helped -

Sleepy~

5 0

3 years ago

Which of the following inequalities is not true?

Galina-37 [17]

The answer is C or the third answer choice because the absolute value of nine is equivalent to the absolute value of nine. The meaning of absolute value is the number of places away from zero on the number line, and both 9 and -9 are equal distance from zero.

6 0

3 years ago

Read 2 more answers

The Earth is one astronomical unit from the sun i.e.1AU= 93 million miles.The angular speed of the Earth traveling around the su

Wewaii [24]

Answer: 0.0667117 * $10^{6}$ miles

Step-by-step explanation: The Earth is one astronomical unit from the sun i.e.1AU= 93 million miles.The angular speed of the Earth traveling around the sun is approximately 0.9863 degrees per day. If we assume the orbit of the Earth is completely circular, what is the linear speed in miles per hour?

Step-by-step explanation: First, a day has 24 hours. Knowing how far the earth travels in day we can say:

0,9863 degrees/day = 0,9863 degrees/(24 hours)

0,9863 degrees/day = 0.0411 degrees/hours

Also, we know the orbit completely circular, so the total distance on, degrees, is 360°. Also this distance can be know with the formula:

Total Circuference = 360° = 2*pi* R

Because Radius is variable for any circle. It is say that:

360° = 2*pi radians

WIth this relation, we can know how many radians are 0.0411 degrees

360/0.0411 = (2*pi)/x

x = (2*pi*0.0411)/(360)

X = 0.00071733032

X is how much the Earth travels per hour in Radians.

Multiplying per Radius:

Earth distance traveled per hour = 0.0667117 * $10^{6}$ miles

8 0

4 years ago

Solve for b 15=3b<br> Please I need help

IgorLugansk [536]

15=3b
15/3 = 3b/b
5= b

Isolate the variable by diving 3 on both sides

6 0

3 years ago

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi

eimsori [14]

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we know the population deviation so 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:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0

4 years ago