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3 years ago

Instructions: Find the missing side lengths. Leave your answers as radicals in simplest form.

Mathematics

1 answer:

ivanzaharov [21]3 years ago

3 0

Answer:

x = 6

y = 3

Step-by-step explanation:

Give triangle is a right triangle.

By applying sine ratio in the given triangle,

sin(60)° = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

$\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{x}$

x = 6

Now we take the cosine ratio in the given triangle,

cos(60)° = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

$\frac{1}{2}=\frac{y}{x}$

y = $\frac{x}{2}$

y = $\frac{6}{2}=3$

Therefore, x = 6 and y = 3 will be the answer.

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H(t) = 210 - 150<br> How do you find the value of t

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Answer:

t = -40

Step-by-step explanation:

the value of t is just the answer. Do not let the H confuse you.

210 - 250 = -40

t = -40

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3 years ago

Maria pulls colored marbles out of a bag one at a time. Her results are shown in the table below.

nevsk [136]

Answer:

Step-by-step explanation:

P(white) = 24/120

set up a proportion based on these results:

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3 years ago

Examination of a large sample of people revealed a relationship between calcium intake and blood pressure. The relationship was

Vika [28.1K]

Answer:

a) The 95% confidence interval for the mean is (-1.171, 11.717).

b) No, there is not enough evidence to support the claim that increasing the amount of calcium in our diet reduce blood pressure (at a 5% significance level).

Step-by-step explanation:

The data for both groups are:

Group 1 (calcium): 7 -4 18 17 -3 -5 1 10 11 -2

Group 2 (placebo): -1 12 -1 -3 3 -5 5 2 -11 -1 -3

The mean and standard deviation for Group 1 is:

$M_1=\dfrac{1}{10}\sum_{i=1}^{10}(7+(-4)+18+17+(-3)+(-5)+1+10+11+(-2))\\\\\\ M_1=\dfrac{50}{10}=5$

$s_1=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{10}(x_i-M)^2}\\\\\\$

$s_1=\sqrt{\dfrac{1}{9}\cdot [(7-(5))^2+(-4-(5))^2+...+(-2-(5))^2]}\\\\\\$

$s_1=\sqrt{\dfrac{1}{9}\cdot [(4)+(81)+(169)+(144)+(64)+(100)+(16)+(25)+(36)+(49)]}\\\\\\s_1=\sqrt{\dfrac{688}{9}}=\sqrt{76.44}\\\\\\s_1=8.743$

The mean and standard deviation for Group 2 is:

$M_2=\dfrac{1}{11}\sum_{i=1}^{11}((-1)+12+(-1)+(-3)+3+(-5)+5+2+(-11)+(-1)+(-3))\\\\\\ M_2=\dfrac{-3}{11}=-0.273$

$s_2=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{11}(x_i-M)^2}\\\\\\$

$s_2=\sqrt{\dfrac{1}{10}\cdot [(-1-(-0.273))^2+(12-(-0.273))^2+...+(-3-(-0.273))^2]}\\\\\\$

$s_2=\sqrt{\dfrac{1}{10}\cdot [(0.53)+(150.62)+...+(115.07)+(0.53)+(7.44)]}\\\\\\s_2=\sqrt{\dfrac{348.182}{10}}=\sqrt{34.82}\\\\\\s_2=5.901$

a) We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The Group 1, of size n1=10 has a mean of 5 and a standard deviation of 8.743.

The Group 2, of size n2=11 has a mean of -0.273 and a standard deviation of 5.901.

The difference between sample means is Md=5.273.

$M_d=M_1-M_2=5-(-0.273)=5.273$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8.743^2}{10}+\dfrac{5.901^2}{11}}\\\\\\s_{M_d}=\sqrt{7.644+3.166}=\sqrt{10.81}=3.288$

The degrees of freedom for this test are:

$df=n_1+n_2-1=10+11-2=19$

The critical t-value for a 95% confidence interval and 19 degrees of freedom is t=1.96.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=1.96 \cdot 3.288=6.444$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 5.273-6.444=-1.171\\\\UL=M_d+t \cdot s_{M_d} = 5.273+6.444=11.717$

The 95% confidence interval for the mean is (-1.171, 11.717).

b) This is a hypothesis test for the difference between populations means.

The claim is that increasing the amount of calcium in our diet reduce blood pressure.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0

The significance level is 0.05.

The sample 1, of size n1=10 has a mean of 5 and a standard deviation of 8.743.

The sample 1, of size n1=11 has a mean of -0.273 and a standard deviation of 5.901.

The difference between sample means is Md=5.273.

$M_d=M_1-M_2=5-(-0.273)=5.273$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{8.743^2}{10}+\dfrac{5.901^2}{11}}\\\\\\s_{M_d}=\sqrt{7.644+3.166}=\sqrt{10.81}=3.288$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{5.273-0}{3.288}=\dfrac{5.273}{3.288}=1.604$