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Kryger [21]
3 years ago
11

You are buying floor tile to cover a room that measures 13 feet by 22 feet. The tile is priced at $19.50 per square yard. How mu

ch will the tile cost?
Mathematics
1 answer:
egoroff_w [7]3 years ago
3 0
1F=1/3Y

(1F)^2=(1/3Y)^2

F^2=1/9Y^2

------------------------

13F x 22F = 286F^2

-------------------------

286F^2 = 286 x 1/9 x Y^2

286F^2 = 286/9 Y^2

-------------------------

286/9 x $19.50 = $619.67

------------------------

Answer:

$619.67
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no, this is not a function

Step-by-step explanation:

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Step-by-step explanation:

\frac{6}{18}=\frac{4}{c}

Solve for c

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3 years ago
A basket holds at most 14 pounds of apples and oranges
Helen [10]
A.

B and C are wrong because x is more than or equal to 3.

D is wrong because y is the weight of oranges and cannot be negative.
7 0
3 years ago
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(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
in a history class, the girl to boy ratio is 9 to 5. if there is a total of 70 students how many boys is there
DedPeter [7]

Answer:

25 boys.

Step-by-step explanation:

To solve this problem, first add the two parts of the ratio. 9 + 5 = 14. Then divide 70 by 14 to get 5. Since the total is 5 times the ratio total, multiply both sides of the ratio by 5. The result is 45 : 25. If you add them together, you get a total of 70. Therefore, there are 25 boys in the class.

7 0
3 years ago
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