What is .7 divided by 21? show your work

A cone-shaped funnel has a diameter of 3.6 in. and a height of 3 in. What is the volume of the funnel? Enter your answer as a de

Sedbober [7]

V=1/3πr^2h
d=3.6 in.
r=1.8 in.
V=1/3×π×1.8^2×3
V≈10.1736 cubic inches. As a result, the volume of the funnel is 10.1736 cubic inches. Hope it help!

3 0

3 years ago

Read 2 more answers

If b is the midpoint of AC, which of the following choices is the value of x?

katrin [286]

If B is the midpoint of AC, that means that the distances from A to B and B to C are the same. The distance from A to B is 8x + 4, and the distance from B to C is 10x - 6, so we can set these two equivalent:

8x + 4 = 10x - 6

Solving for x:

4 = 2x - 6 (subtract 8x from both sides)
10 = 2x (add 6 to both sides)
5 = x (divide both sides by 2)

So, x = 5, which is the second option given.

8 0

3 years ago

If a building was placed at coordinates (-5,5), what would be the new coordinates if rotated 270° counterclockwise around the or

Brums [2.3K]

The rule for a 270° counterclockwise rotation is (x, y) ---> (y, -x).

So if you rotated (-5,5) 270° counterclockwise around the origin the new coordinates would be (-5,-5).

Hope this Helps!!

6 0

3 years ago

Read 2 more answers

Matthew purchased a desk that was on

natulia [17]

Answer:

ANSWER :

285.12

discount amount =45/00 x 480=45 x 48/10

=216.0

price of discount =480-216=264

salex tex 8/100 x 264. =21.12

total many of spend 264+21.12=285.12

4 0

3 years ago

Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

Vesnalui [34]

Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ 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 check if the mean is less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and we don't 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:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

$df=n-1=40-1=39$

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

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0

3 years ago