All categories

2 years ago

The price of an item has been reduced by 30%. The original price was $20. What is the price of the item now?

Mathematics

2 answers:

Ludmilka [50]2 years ago

7 0

Answer:

$14

Step-by-step explanation:

multiply .30 times 20 to get $6, subtract that from the orginal price which is $20

Ugo [173]2 years ago

4 0

$6 because if you do 20 times 0.3 (30%) then it’ll give u $6

You might be interested in

Solve the following system of equations.<br> 5x+2y =-6<br> 3x-2y = 22

katen-ka-za [31]

Answer:

5x+2y =-6 ........1

3x-2y = 22 ........2

Add both Equations

we get

8x = 16

Divide both sides by 8

x = 2

Substitute x = 2 into any of the equations

3(2) - 2y = 22

-2y = 22 - 6

- 2y = 18

Divide both sides by -2

y = - 9

x = 2 y = - 9

Hope this helps

6 0

3 years ago

Solve for c.

Vanyuwa [196]

Answer:

Step-by-step explanation:

solve for c by simplifying both sides of the equation then isolate the variable

6 0

2 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

2 years ago

What is the distance between points a and e

kozerog [31]

Please post the question with an attachment of what you want to be answered next time.

7 0

3 years ago

Derivative of Y=cos^2(3x)

Black_prince [1.1K]

The derivative is -6sin(3x)

4 0

3 years ago