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Sladkaya [172]
3 years ago
10

An airline charges $150 for a ticket and $20 for each bag. Which equation shows the relationship between the number of bag, b ,

and the total cost, c , for flying with this airline?
Mathematics
1 answer:
zhenek [66]3 years ago
5 0

Answer:

c = 150 + 20b

Step-by-step explanation:

Given:

Ticket charges = $150

Each bag carry cost = $20

Find:

Relationship between bag(b) and cost(c)

Computation:

⇒ Total cost = Ticket charges + $20(number of bags)

⇒ c = 150 + 20(b)

⇒ c = 150 + 20b

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What is the following product
Genrish500 [490]

For this case we must find the product of the following expression:

\sqrt [3] {5} * \sqrt {2}

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5 ^ {\frac {1} {3}} * 2 ^ {\frac {1} {2}} =

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ANswer:

Option b

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What is StartAbsoluteValue 8 EndAbsoluteValue? –16 –8 8 16 help plz
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Read 2 more answers
(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
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