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3 years ago

Find the value of x.

Mathematics

1 answer:

son4ous [18]3 years ago

8 0

Answer:

18.0

Step-by-step explanation:

==>Given:

Triangle with sides, 16, 30, and x, and a measure of an angle corresponding to x = 30°

==>Required:

Value of x to the nearest tenth

==>Solution:

Using the Cosine rule: c² = a² + b² - 2abcos(C)

Let c = x,

a = 16

b = 30

C = 30°

Thus,

c² = 16² + 30² - 2*16*30*cos 30°

c² = 256 + 900 - 960 * 0.8660

c² = 1,156 - 831.36

c² = 324.64

c = √324.64

c = 18.017769

x ≈ 18.0 (rounded to nearest tenth)

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Consider the equation 2/5y = 20. Tell whether each statement is true or false

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Answer:

A. True
B. False
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Step-by-step explanation:

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2 years ago

adults say that they have cheated on a test or exam before. You randomly select six adults. Find the probability that the number

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Answer: a. 1/6. b. 2/3. c. 5/6

Step-by-step explanation:

Number of adults randomly selected= 6

a. Finding the probability of adults that have cheated on an examination before is exactly 4.

The numbers are 1, 2, 3, 4, 5 and 6.

Probability of chosen exactly 4 will be 1/6 since 4 can only occur once.

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3 years ago

(4,?), is on the line below. Find the other half of the coordinate. Y= - 3/4x + 3

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3 years ago

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.