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2 years ago

5

A home builder needs to cover the rectangular floor of a closet with a carpet the closet is 2 yards long and 2/3 of a yard wide

how much carpet will the home builder need to cover the closet?

1 answer:

Crank2 years ago

7 0

Your Answer is 4 Yards

Step-by-step explanation:

2/3 yards is 2 feet

2 yards is 6 feet

2 times 6 feet is 12 foot e.g. (2 x 6 = 12)

12 feet is 4 yards.

Formula (L x W = A)

L = length

W = Width

A = Area

<em>Hope this helps! Please let me know if you need more help or think my answer is incorrect. Brainliest would be MUCH appreciated. Have a wonderful day!</em>

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A piece of paper is randomly selected from a stack that contains 15 small, 44 large, and 20 extra large pieces.

yKpoI14uk [10]

Answer: The true statement is (A) P(extra large)≈0.25.

Step-by-step explanation:

Given that a piece of paper is randomly selected from a stack that contains 15 small, 44 large, and 20 extra large pieces.

We are to select the true statement.

Let S be the sample space for the experiment of choosing a piece.

Then, n(S) = 15 + 44 + 20 = 79.

Let A, B and C denote the events of choosing a small piece, a large piece and an extra large piece respectively.

So, n(A) = 15, n(B) = 44 and n(C) = 20.

Therefore, we have

Thus, the true statement is P(extra large)≈0.25.

Option (A) is correct

3 0

3 years ago

B. There are 4 quarts in a gallon and 4 cups in<br> a quart. How many cups are in one serving?

Brrunno [24]

Answer:

16 cups

Step-by-step explanation:

4 0

3 years ago

An isosceles right triangle has a hypotenuse of 16mm what would the area of the triangle be

olga2289 [7]

Answer: 64 square mm

Step-by-step explanation:

6 0

2 years ago

Order these fractions from least to greatest 3/4 2/5 5/8 1/2

allsm [11]

Answer:

2/5 <1/2 <5/8<3/4

Step-by-step explanation:

2/5 ⇒40%

1/2 ⇒50%

3/4⇒75/%

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6 0

4 years ago

Read 2 more answers

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.

8 0

3 years ago