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3 years ago

True or false the foci of an ellipes can be outside the ellipes

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

6 0

Answer:

false

Step-by-step explanation:

foci are usually in the ellipse. that foci is where the ellipse is drawn from

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Solve the following:<br> 3/4x=1/8<br> x/5=2.3<br> x/3.2=4.5

antiseptic1488 [7]

Step-by-step explanation:

mahirap yung tanong at pinapasagot mo

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3 years ago

HELPPP ILL DO ANYTHING HELPPPP

katovenus [111]

Answer:

-2

Step-by-step explanation:

hi! to find f(-3) on this graph, go to the x-value -3 and find the y-value there. when we look at -3 on the x-axis, we go down and see that there is a point at (-3,-2). therefore, f(-3)=-2.

3 0

3 years ago

PLEASE HELP ASAP!!

Art [367]

Answer:

See explanation

Step-by-step explanation:

A. P(top)=top outcomes/all kinds of outcomes=4/30=2/15=13.33333333333...%

P(bottom)=bottom outcomes/ all kinds of outcomes=1/30=3.33333333....%

P(side)=25/30=5/6=83.33333333333...%

B. No. If were equally likely, the probabilities for A would have been roughly the same. It seems like the side event is more probable.

7 0

3 years ago

A recycling bin is in the shape of a right rectangular prism. The bin is 12 meters long, 512 meters wide, and 612 meters tall.

devlian [24]

To find the volume of the recycling bin, you will use the formula for finding volume of a rectangular prism.

V = lwh
V = 12 x 512 x 612
V = 3760128 cubic meters

The answer is not given, but based in these numbers this is correct.

If it is 12 m x 5 m x 6 than the answer is 360 cubic meters

6 0

3 years ago

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190$

The 95% confidence interval for the mean difference is (490, 1190).

7 0

3 years ago