All categories

3 years ago

Find the center, vertices, and foci for the ellipse 4x2 + 25y2 = 100.

Mathematics

1 answer:

navik [9.2K]3 years ago

6 0

Answer: C

Step-by-step explanation:

Center: (0,0)

Vertices: (+/-5,0)

Foci: (+/-4.6,0)

You might be interested in

(15 – 5) •2<br> ill give brainlist

IRISSAK [1]

Answer:

Step-by-step explanation:15-5=10 10 x 2 =20

8 0

3 years ago

Read 2 more answers

In long division what is 601 divided by 3

horrorfan [7]

Who knows cause I don't

4 0

3 years ago

I dont understand this but maybe some one will and they will help me. The quotient of a number and -7 decreased by 2 is ten find

givi [52]

It might be nineteen because nineteen minus seven is twelve. twelve minus two is ten.

8 0

3 years ago

A jar has 3 red marbles and 9 blue ones. If you randomly pick a marble without replacing it and then select another, what is the

GaryK [48]

The probability of picking two blue would be 2/9 because you would use the 2 out of the 9

5 0

3 years ago

Read 2 more answers

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago