Ask question

Ask question

All categories

3 years ago

14

The length of the sides of a triangular garden are 10m 11m 13m.Find the measure of the angle opposite the longest side to the ne

arest tenth of a degree

1 answer:

Kaylis [27]3 years ago

3 0

C^2=a^2+b^2-2abcosC
13^2=10^2+11^2-2×10×11×cos(°)
169=100+121-220cos(°)
220cos(°)=100+121-169
220cos(°)=52
cos(°)= 13/55
°=cos-1(13/55)
°=76.328°

You might be interested in

Please help DO NOT IGNORE DO NOT IGNORE

lisov135 [29]

Answer:

24.997 cm

Step-by-step explanation:

Find the diameter:

7(2) = 14 cm

Find the circumference:

C = $\pi d$

C = $\pi (14)$

C = 43.988

Divide the circumference by 4:

43.988/4 = 10.997

(This is the length of the curved side)

Add length to the other 2 sides:

7 + 7 + 10.997 = 24.997

3 0

3 years ago

What is the answer to the whole thing? I'm having trouble with my homework.

notsponge [240]

Y=8,4,2,-2 linear and in that order

8 0

3 years ago

Read 2 more answers

I can’t figure this out

patriot [66]

Answer:

$m \leq 4.3$

Step-by-step explanation:

We need to get m on one side all by itself. To achieve this, let's subtract both sides by 1.2 to get $m \leq 5.5-1.2 \implies m \leq 4.3$ .

7 0

3 years ago

A sequence of 1 million iid symbols(+1 and +2), Xi, are transmitted through a channel and summed to produce a new random variabl

Zigmanuir [339]

Answer:

E(w) = 1600000

v(w) = 240000

Step-by-step explanation:

given data

sequence = 1 million iid (+1 and +2)

probability of transmitting a +1 = 0.4

solution

sequence will be here as

P{Xi = k } = 0.4 for k = +1

0.6 for k = +2

and define is

x1 + x2 + ................ + X1000000

so for expected value for W

E(w) = E( x1 + x2 + ................ + X1000000 ) ......................1

as per the linear probability of expectation

E(w) = 1000000 ( 0.4 × 1 + 0.6 × 2)

E(w) = 1600000

and

for variance of W

v(w) = V ( x1 + x2 + ................ + X1000000 ) ..........................2

v(w) = V x1 + V x2 + ................ + V X1000000

here also same as that xi are i.e d so cov(xi, xj ) = 0 and i ≠ j

so

v(w) = 1000000 ( v(x) )

v(w) = 1000000 ( 0.24)

v(w) = 240000

8 0

3 years ago

In 2002, the mean age of an inmate on death row was 40.7 years with a standard deviation of 9.6 years according to the U.S. Depa

marissa [1.9K]

Answer:

The <em>95% confidence interval</em> for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Step-by-step explanation:

The <em>confidence interval</em> of the mean is given by the next formula:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$ [1]

We already know (according to the U.S. Department of Justice):

The (population) standard deviation for this case (mean age of an inmate on death row) has a standard deviation of 9.6 years ( $\\ \sigma = 9.6$ years).
The number of observations for the sample taken is $\\ n = 32$ .
The sample mean, $\\ \overline{x} = 38.9$ years.

For $\\ z_{1-\frac{\alpha}{2}}$ , we have that $\\ \alpha = 0.05$ . That is, the <em>level of significance</em> $\\ \alpha$ is 1 - 0.95 = 0.05. In this case, then, we have that the <em>z-score</em> corresponding to this case is:

$\\ z_{1-\frac{\alpha}{2}} = z_{1-\frac{0.05}{2}} = z_{1-0.025} = z_{0.975}$

Consulting a cumulative <em>standard normal table</em>, available on the Internet or in Statistics books, to find the z-score associated to the probability of, $\\ P(z$ , we have that $\\ z = 1.96$ .

Notice that we supposed that the sample is from a population that follows a <em>normal distribution</em>. However, we also have a value for n > 30, and we already know that for this result the sampling distribution for the sample means follows, approximately, a normal distribution with mean, $\\ \mu$ , and standard deviation, $\\ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$ .

Having all this information, we can proceed to answer the question.

Constructing the 95% confidence interval for the current mean age of death-row inmates

To construct the 95% confidence interval, we already know that this interval is given by [1]:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

That is, we have:

$\\ \overline{x} = 38.9$ years.

$\\ z_{1-\frac{\alpha}{2}} = 1.96$

$\\ \sigma = 9.6$ years.

$\\ n = 32$

Then

$\\ 38.9 \pm 1.96*\frac{9.6}{\sqrt{32}}$

$\\ 38.9 \pm 1.96*\frac{9.6}{5.656854}$

$\\ 38.9 \pm 1.96*1.697056$

$\\ 38.9 \pm 3.326229$

Therefore, the Upper and Lower limits of the interval are:

Upper limit:

$\\ 38.9 + 3.326229$

$\\ 42.226229 \approx 42.23$ years.

Lower limit:

$\\ 38.9 - 3.326229$

$\\ 35.573771 \approx 35.57$ years.

In sum, the 95% confidence interval for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Notice that the "mean age of an inmate on death row was 40.7 years in 2002", and this value is between the limits of the 95% confidence interval obtained. So, according to the random sample under study, it seems that this mean age has not changed.

7 0

3 years ago