All categories

2 years ago

Determine whether the relation defines y as a function of x. Give the domain.

Mathematics

1 answer:

Schach [20]2 years ago

5 0

Answer:

sorry i need things to make a question

Step-by-step explanation:

its soooooooooooooooo hard

You might be interested in

Find the y-intercept for the parabola defined by<br> this equation:<br> y=2x2+x−1

Natalija [7]

Answer:

(0, -1)

Step-by-step explanation:

The c value of the equation represents the y-intercept. c=-1 so the y-intercept must be (0,-1)

7 0

2 years ago

Cesar made a rectangular banner that is 4 feet by 3 feet. he wants to make a triangular banner that has the same area as the oth

dimulka [17.4K]

6 Because 4*3=12 and 4*6/2=12

6 0

3 years ago

Need help please on Pre Calculus!!! Directions: verify each identity

Juliette [100K]

You first simplify

tan(x) = cos(x)/sin(x) and cot(x) = sin(x)/cos(x)

Then you plug that in and then you simplify even more

3 0

3 years ago

How do I prove this?

Vadim26 [7]

Incorrect info please upload a correct photo of the problem

6 0

3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

3 years ago