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3 years ago

Is the following expression shown the exact answer? 144π - 216√3 true or false

1 answer:

6 0

$144\pi - 216\sqrt{3} \\\\
144 = 2^4 \times 3^2\\\\
216 = 2^3 \times 3^3\\\\
\Longrightarrow~~\text{Common factor } = 2^3 \times 3^2 = 8 \times 9 = 72\\\\
\Longrightarrow~~
144\pi - 216\sqrt{3} = \boxed{72\Big(2\pi - 3\sqrt{3} \Big)}$

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Nesterboy [21]

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3 years ago

What is the quotient of 2 1/9÷3 4/5

Karolina [17]

Answer:

$\frac{5}{9}$

Step-by-step explanation:

$2 \frac{1}{9} \div 3 \frac{4}{5} \\ \\ = \frac{2 \times 9 + 1}{9} \div \frac{3 \times 5 + 4}{5} \\ \\ = \frac{18 + 1}{9} \div \frac{15 + 4}{5}\\ \\ = \frac{19}{9} \div \frac{19}{5}\\ \\ = \frac{19}{9} \times \frac{5}{19} \\ \\ = \frac{5}{9}$

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3 years ago

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2 years ago

Convert a recipe that serves 6 and calls for 3/4 cups of sugar to one that serves 20.

djyliett [7]

Answer:

5/2cups

Step-by-step explanation:

If a recipe serves 6 and calls for 3/4cups, arecipe that serves 20, will call for (20×3/4)/6 cups. Mathematically;

6recipe = 3/4cups

20recipe = x

Cross multiplying we have;

6x = 20×3/4

6x = 15

x = 15/6

x = 5/2

This means a recipe that serves 20 will call for 5/2cups

3 0

3 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago