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frosja888 [35]
1 year ago
10

The number of hamburger sales per week at a fast food restaurant is 350. The number of weekly sales goes down by 30. 0 hamburger

s per week when the price of a hamburger is increased by 12%. Find the relative change in weekly sales
Mathematics
1 answer:
Airida [17]1 year ago
8 0

The relative change in weekly sales is  mathematically given as

relative change= -12.857%

This is further explained below.

<h3>What is the relative change in weekly sales?</h3>

This Sale per week after the price decreases

=350-45  

=305

Generally, the equation for relative change is  mathematically given as

\left(\frac{\text { final Value - Intial value }}{\text { Intial value }}\right) \times 100 \%$

Hore final value =305 week

Initial Value =350 week

Relative Change=\left(\frac{305-350}{350}\right) \times 100

&=\frac{-45}{350} \times 100 \\

=-12.857%

Read more about  relative change

brainly.com/question/22278222

#SPJ1

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Hey u plz help me me need help
cestrela7 [59]
The answer is b lol
4 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
2 years ago
The question is above in the picture​
anzhelika [568]
Yea his answer is right ^^
6 0
3 years ago
3х - 3 = -12<br> What is the solution to this equation?
Lapatulllka [165]
3x-3=-12

3x=-9

X=-3


I am in calculus. Algebra is easy
5 0
2 years ago
Our club shirts are available in two sizes: small shirts coat $27 each and large shirts cost $31 each. Last week a total of 14 s
DanielleElmas [232]

small = 9

large = 5

hope this helps

7 0
3 years ago
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