First, determine the area of the floor of the new room by multiplying the dimensions in feet.
Area = (12 ft)(15.5 ft) = 186 ft²
Then, we convert the calculated area to yards
(186 ft²)(1 yd/ 3 ft)² = 20.67 yd²
Given that the supplier only delivered 20 yd², approximately 0.67 yd² of your floor will not have carpet.

6 0

3 years ago

Find the quotient. Do not round your answer

adoni [48]

5.4945 is the answer not rounded

7 0

3 years ago

Read 2 more answers

Quarters are currently minted with weights normally distributed and having a standard deviation of 0.065. New equipment is being

Ilya [14]

Answer:

Yes, the new equipment appear to be effective in reducing the variation of weights.

Step-by-step explanation:

We are given that Quarters are currently minted with weights normally distributed and having a standard deviation of 0.065.

A simple random sample of 25 quarters is obtained from those manufactured with the new equipment, and this sample has a standard deviation of 0.047.

Let $\sigma$ = standard deviation of weights of new equipment.

SO, Null Hypothesis, $H_0$ : $\sigma \geq$ 0.065 {means that the new equipment have weights with a standard deviation more than or equal to 0.065}

Alternate Hypothesis, $H_A$ : $\sigma$ < 0.065 {means that the new equipment have weights with a standard deviation less than 0.065}

The test statistics that would be used here One-sample chi-square test statistics;

T.S. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2}__n_-_1$

where, s = sample standard deviation = 0.047

n = sample of quarters = 25

So, the test statistics = $\frac{(25-1)\times 0.047^{2} }{0.065^{2} }$ ~ $\chi^{2}__2_4$

= 12.55

The value of chi-square test statistics is 12.55.

Now, at 0.05 significance level the chi-square table gives critical value of 13.85 at 24 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of chi-square as 12.55 < 13.85, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the new equipment have weights with a standard deviation less than 0.065.

5 0

3 years ago