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1 year ago

Help me please!!! 10 points!!

Mathematics

2 answers:

hammer [34]1 year ago

8 0

$\qquad\qquad\huge\underline{{\sf Answer}}$

Let's solve for x ~

We can apply Trigonometric ratios here, since the given triangle is a right angled triangle

so,

$\qquad \sf \dashrightarrow \: \sin(30) = \dfrac{ \sqrt{3} }{x}$

$\qquad \sf \dashrightarrow \: \dfrac{1}{2} = \dfrac{ \sqrt{3} }{x}$

$\qquad \sf \dashrightarrow \:x = 2 \times \sqrt{3}$

$\qquad \sf \dashrightarrow \:x = 2 \sqrt{3}$

i hope you you understood

Allisa [31]1 year ago

4 0

Answer:

2√3

Step-by-step explanation:

The shown triangle is a 30-60-90 special right triangle due to the right angle and the thirty degree angle(the other angle has to be 60 degrees).

The hypotenuse of a 30-60-90 triangle is twice the leg opposite the 30 angle. The leg opposite the 30 degree angle is √3, so x would be 2√3.

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zalisa [80]

Answer:

129.5 Square feet is the correct answer.

Step-by-step explanation:

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3 years ago

P = 2L + 2W; for L NEED HELP PLEASE

Orlov [11]

L=(P-2W)/2 HOPE I HELPED YA :)

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2 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.