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3 years ago

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12. Nitrogen is a gas at certain temperatures under normal conditions. These temperatures (1), in degrees

1 answer:

frozen [14]3 years ago

3 0

The final answer is C

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Find the value of x in the triangle. SU RV

kodGreya [7K]

Answer:

x = 8

Step-by-step explanation:

Δ TSU and Δ TRV are similar , thus the ratios of corresponding sides are equal, that is

$\frac{TS}{TR}$ = $\frac{TU}{TV}$ , substitute values

$\frac{x}{x+16}$ = $\frac{5}{15}$ = $\frac{1}{3}$ ( cross- multiply )

3x = x + 16 ( subtract x from both sides )

2x = 16 ( divide both sides by 2 )

x = 8

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3 years ago

the sum of the first n terms of a sequence is 20 - 10 / 2 ^ n - 1 find the sum of the first five terms

Travka [436]

Answer:

$S_5=\dfrac{170}{9}$

Step-by-step explanation:

Given that,

The sum of first n term of a sequence is $S_n=20-\dfrac{10}{2n-1}$

We need to find the sum of the first 5 terms.

Put n = 5,

$S_5=20-\dfrac{10}{2(5)-1}\\\\=20-\dfrac{10}{10-1}\\\\=20-\dfrac{10}{9}\\\\=\dfrac{170}{9}$

So, the sum of first 5 terms is equal to $\dfrac{170}{9}$ .

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2 years ago

Of 960 pages 384 had an add. What's the probability it contains an add

kramer

384/960
Thus, answer=2/5

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3 years ago

18.194 rounded to the nearest hundredths

ehidna [41]

18.19
-you're welcome

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3 years ago

Read 2 more answers

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