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3 years ago

7

Can someone help me with this problem ASAP please.

1 answer:

harina [27]3 years ago

3 0

None are correct because angle A is 40 degrees

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Select 1 answer choices: sizes averages distances areas perimeters

hammer [34]

Answer:

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Step-by-step explanation:

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7 0

3 years ago

In 2009, a town's population was about 9,000 residents. In 2014, the population was about 12,500 residents. Which linear model r

jek_recluse [69]

Answer:

The linear function that discribes the size of the population in function of the t in years is p = 700t - 1,397,300

Step-by-step explanation:

A linear function is defined by a line, so in order to determine the linear function we can use the two points that were given to us to create a line equation and use that as our linear function. The points given to us were (2009; 9000) and (2014; 12500), in this case the year is our value of "x" and the size of the population is our value of "y". The first step is to find the slope of the line which is given by:

m = (y2 - y1)/(x2 - x1)

m = (12500 -9000)/(2014 - 2009) = 3500/5 = 700

Then we can use the slope and the first point to build the equation:

p - 9000 = 700*(t - 2009)

p = 700t - 1406300 + 9000

p = 700t - 1397300

5 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

Helpphelpphelpphelpphelpp<br> helpphelpphelpphelpphelpphelpphelpphelpphelpphelpphelpphelpphelpp

umka2103 [35]

Answer:

wait what?

Step-by-step explanation:

7 0

3 years ago

Express the sum of 3x^2+5x-6 and -x^2+3×+9?

djverab [1.8K]

I hope this helps you

6 0

3 years ago