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3 years ago

In the 30-60-90 triangle below, side s has a length of _ and side q has a length of _.

Mathematics

1 answer:

gregori [183]3 years ago

4 0

Answer: F

Step-by-step explanation:

For a 30-60-90 triangle, we know that the hypotenuse is 2x. Since we know the hypotenuse is 10, we can solve for x.

2x=10

x=5

Now that we know x is 5, we can use this to solve for s and q. The side across from 30° is just x. Since we know x, s is 5.

The side across from 60° is x√3. Since we know what x is, we can just plug in. q is 5√3.

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sdas [7]

Answer:

Step-by-step explanation:

Hello!

The variables of interest are:

X₁: Number of cases of polio observed in kids that received the placebo vaccine.

n₁= 201299 total children studied

x₁= 110 cases observed

X₂: Number of cases of polio observed in kids that received the experimental vaccine.

n₂= 200745 total children studied

x₂= 33 cases observed

These two variables have a binomial distribution. The parameters of interest, the ones to compare, are the population proportions: p₁ vs p₂

You have to test if the population proportions of children who contracted polio in both groups are different: p₂ ≠ p₁

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.05

$Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }$

Sample proportion placebo p'₁= x₁/n₁= 110/201299= 0.0005

Sample proportion vaccine p'₂= x₂/n₂= 33/200745= 0.0002

Pooled sample proportion p'= (x₁+x₂)/(n₁+n₂)= (110+33)/(201299+200745)= 0.0004

$Z_{H_0}= \frac{(0.0002-0.0005)-0}{\sqrt{0.0004[\frac{1}{201299} +\frac{1}{200745} ]} }= -4.76$

This test is two-tailed, using the critical value approach, you have to determine two critical values:

$Z_{\alpha/2}= Z_{0.025}= -1.96$

$Z_{1-\alpha /2}= Z_{0.975}= 1.96$

Then if $Z_{H_0}$ ≤ -1.96 or if $Z_{H_0}$ ≥ 1.96, the decision is to reject the null hypothesis.

If -1.96 < $Z_{H_0}$ < 1.96, the decision is to not reject the null hypothesis.

⇒ $Z_{H_0}$ = -4.76, the decision is to reject the null hypothesis.

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.01

$Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }$

The value of $Z_{H_0}$ = -4.76 doesn't change, since we are working with the same samples.

The only thing that changes alongside with the level of significance is the rejection region:

$Z_{\alpha /2}= Z_{0.005}= -2.576$

$Z_{1-\alpha /2}= Z_{0.995}= 2.576$

Then if $Z_{H_0}$ ≤ -2.576or if $Z_{H_0}$ ≥ 2.576, the decision is to reject the null hypothesis.

If -2.576< $Z_{H_0}$ < 2.576, the decision is to not reject the null hypothesis.

⇒ $Z_{H_0}$ = -4.76, the decision is to reject the null hypothesis.

Remember the level of significance (probability of committing type I error) is the probability of rejecting a true null hypothesis. This means that the smaller this value is, the fewer chances you have of discarding the true null hypothesis. But as you know, you cannot just reduce this value to zero because, the smaller α is, the bigger β (probability of committing type II error) becomes.

Rejecting the null hypothesis using different values of α means that there is a high chance that you reached a correct decision (rejecting a false null hypothesis)

I hope this helps!

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3 years ago