All categories

3 years ago

Compute the number of vinyl tiles, measuring 28 inches on each side, needed to tile a kitchen measuring 102 ft. by 18 ft.

Mathematics

2 answers:

NikAS [45]3 years ago

8 0

Answer:

338

Step-by-step explanation:

math

ratelena [41]3 years ago

6 0

Answer:

338 tiles

Step-by-step explanation:

First, convert the measurements of the kitchen from feet to inches by multiplying each measurement by 12.

102 * 12 = 1224 18 * 12 = 216

Now multiply these answers together.

1224 * 216 = 264,384 square inches

Now find the area of the vinyl tiles.

28 * 28 = 784 square inches

Now find how many 784 square inch tiles are needed to cover a 264,384 square inch kitchen.

264,384 / 784 = 337.224

Since you can't buy part of a tile, round your answer up.

338 tiles

You might be interested in

How many ways can 5 books be arranged on a shelf

Zielflug [23.3K]

There are very many ways to this!
We can have horizontal, vertical, some can be laying down wile others are standing up, some could be standing up and the others layed down on top of them etc.
but in this case it would be 120 ways.

3 0

3 years ago

What is the lateral area of a right cylinder if the radius of the base is 12 mm and the altitude of the cylinder is 5 times the

schepotkina [342]

Lateral Area of a cylinder is ; 2πR.H

R=12 mm and H= 15 mm

Then LATERAL area = 2π(12)(15) =48π ≈ 150 mm²

5 0

3 years ago

he American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 45 w

Sonja [21]

Answer:

The 85% confidence interval is (0.2187, 0.2613).

Step-by-step explanation:

We have a large sample of size n = 830 Americans over 45. We have observed that 631 people don't smoke. Let p be the population proportion of Americans over 45 who smoke. An point estimate of p is $\hat{p} = 199/830 = 0.24$ and an estimate of the standard deviation of $\hat{p}$ is $\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{(0.24)(0.76)/830}$ = 0.0148. Therefore, because we have a large sample, the 85% confidence interval for the population proportion is given by $\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n} = 0.24\pm z_{0.15/2}0.0148 = 0.24\pm z_{0.075}0.0148$ where $z_{0.075}$ is the 7.5th quantile of the standard normal distribution, i.e., -1.4395. Therefore the 85% confidence interval is $0.24\pm (1.4395)(0.0148)$ or equivalently (0.2187, 0.2613).

7 0

3 years ago

Can you please help me out

Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

$\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}$

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,

$\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}$ $\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}$

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

$\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}$

Hence, the best option is G.

5 0

1 year ago

Help <br><br> What is the standard form of this????

Reil [10]

The answer is x-3y=-12 use math-way for this

4 0

4 years ago