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2 years ago

Cos B= 4/5, a=10, C=90, find the hypotenuse of the triangle

Mathematics

1 answer:

vlabodo [156]2 years ago

4 0

The cosine of B = 4/5 so the adjacent side divided by the hypotenuse = .8
So 10/.8 = 12.5 the length of the hypotenuse

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QUICK! A. –27 B. –3 C. 3 D. 27

Alex

Answer:

C. 3

Step-by-step explanation:

-9 * -1/3

= -9/ -3

A negative divided by a negative is a positive

9/3

7 0

3 years ago

2) The distance between two villages is 0.625 km.

Elena-2011 [213]

Answer:

12.5 cm

Step-by-step explanation:

0.625km=62500cm. The scale factor is 1:5000. So the distance on the map will be 62500/5000=12.5 cm

6 0

2 years ago

Read 2 more answers

What is the value of x?

Fittoniya [83]

Let the 2 equal lengths be represented by 2p. Then you can write the equation of ratios

p 2p
---- = -------
16 x+4

which simplifies to

1 2
----- = ------
16 x+4

then x+4 = 32, and x = 28 (answer)

8 0

2 years ago

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a significance level of 0.05 and 18 - 1 = 17 df.

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, hence the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0

2 years ago

Name the force that is responsible for the upward movement of water in xylem tissues

GenaCL600 [577]

Answer:

Root Pressure is the answer

6 0

2 years ago