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3 years ago

What is a possible value of sin(theta) when cos2(theta)=0.73?

1 answer:

8 0

We use the following trigonometric identity to answer the question:
sin^ 2 (theta) + cos^ 2 (theta) = 1
We substitute the values:
sin ^ 2 (theta) + 0.73 = 1
We cleared:
sin ^ 2 (theta) = 1-0.73
sin ^ 2 (theta) = 0.27
Answer:
a possible value of sin ^ 2 (theta) is:
c.0.26 (nearest option)

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What is the range of the set of data : 13, 16, 10, 15, 6, 6

aniked [119]

The range is the difference between the largest and smallest number. So for those numbers the lowest value is 6 and the highest is 16. So 16 - 6 = 10

So the range is 10

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2 years ago

Write and solve an equation to find the number n.

garik1379 [7]

The unknown number is n. So n is the unknown variable in the equation.
In order to write an equation in which n is included we must know some dependence of n with other values.
We know that <span>the difference of three times a number and 44 is −19. So, we can write:
3*n-44=-19
3*n=-19+44
3*n=25
n=25/3

</span>

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2 years ago

What are the slope and the y-intercept of the graph of the linear function shown on the grid?

poizon [28]

Y-intercept looks like -1.5 and slope is -1/4

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2 years ago

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who w

dimulka [17.4K]

Answer:

99% CI:

$-4.637\leq\mu_v-\mu_o\leq 3.737$

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}$

The MSE can be calculated as

$MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65$

The weighted sample size nh can be calculated as

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16$

The standard deviation then becomes

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623$

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

$\Delta M=M_v-M_o=5.10-5.55=-0.45$

Then the confidence interval can be constructed as

$\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737$

8 0

2 years ago

Carlos wrote three numbers between 0.33 and 0.34. What numbers could he have written?

kiruha [24]

There is an infinite number of numbers between 0.33 and 0.34
for example, 0.331, 0.332, 0.3321213323, 0.3333333, 0.3336778, etc.
There should be more to the question.

5 0

2 years ago