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3 years ago

Which expressions represent the distances from the point (x,y) to the focus and the directrix?

Mathematics

2 answers:

Amanda [17]3 years ago

6 0

The answer is

The distance from the point (x,y) to the focus, (−2,3), is (x + 2)^2+(y − 3)^2. The distance from the point (x,y) to the directrix, y=7, is |y−7|.

LuckyWell [14K]3 years ago

4 0

Answer:

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Step-by-step explanation:sry i ran out of ponts from one queshten

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HELP ME PLEASE! Vector's map has the scaling missing.he knows from the lake to the youth center is 8 miles. on the map, they are

Naddik [55]

Hey there!

Since 2 goes into 8, you would simplify. You would get 1 inch and 4 miles. Next, set up a ratio. The ratio is 1 inch: 4 miles. So for every inch on the map, there is 4 miles.

I hope this helps!

7 0

3 years ago

An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co

levacccp [35]

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)

2. $4.7048 \leq \sigma^2 \leq 16.1961$

3. $t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

$t_{crit}=1.753$

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. $t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

$\chi^2 =24.9958$

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

$\bar X=26.31$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that $t_{\alpha/2}=2.13$

Now we have everything in order to replace into formula (1):

$26.31-2.13\frac{2.8}{\sqrt{16}}=24.819$

$26.31+2.13\frac{2.8}{\sqrt{16}}=27.801$

So on this case the 95% confidence interval would be given by (24.8190;27.8010)

Part 2

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

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

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

$\chi^2_{\alpha/2}=24.996$

$\chi^2_{1- \alpha/2}=7.261$

And replacing into the formula for the interval we got:

$\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}$

$4.7048 \leq \sigma^2 \leq 16.1961$

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:

Null hypothesis: $\mu \leq 25$

Alternative hypothesis: $\mu > 25$

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)

Calculate the statistic

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

$t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

Critical value

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got $t_{crit}=1.753$

Conclusion

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: $\sigma \leq 2.3$

H1: $\sigma >2.3$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be $\chi^2 =24.9958$

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

7 0

3 years ago

Each carton 12 eggs. There are 2 full cartons in the refrigerator. Margot uses 3 eggs to make a quiche. How many eggs are left.

3241004551 [841]

Answer:

Answer: There are 21 eggs left.

Step-by-step explanation:

One carton has 12 eggs.

2 cartons are two times one carton, so two cartons have 2 times 12 eggs.

2 * 12 = 24

There are 24 eggs in two cartons.

Margot uses 3 eggs. We subtract 3 eggs from 24 eggs.

24 - 3 = 21

Answer: There are 21 eggs left.

6 0

3 years ago

A can in the shape of a cylinder has a diameter of 10 cm and a height of 6 cm. how much wrapping paper is needed to cover the cy

Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

To determine the amount of wrapping paper needed to cover the can, we would determine the total surface area of the cylindrical can. The formula for determining the total surface area of a cylinder is expressed as

Total surface area = 2πr² + 2πrh

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant whose value is 3.14

From the information given,

Diameter = 10 cm

Radius = diameter/2 = 10/2

Radius = 5 cm

Height = 6 cm

Therefore,

Total surface area = (2 × 3.14 × 5²) + (2 × 3.14 × 5 × 6)

= 157 + 188.4 = 345.4 cm²

345 cm² of wrapping paper is needed to cover the cylinder.

2) Volume of cylinder = πr²h

Volume of cone = 1/3πr²h

3 × Volume of cone = πr²h

The volume of the cylinder is 3 times greater than the volume of the cone. Therefore, the volume of the cone is 1/3 times lesser than the volume if the cylinder.

5 0

3 years ago

Solve for x:<br> 4x/5=20

alexdok [17]

The correct response is X=25

3 0

3 years ago