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enot [183]
3 years ago
15

Selena and Drake are evaluating the expression , when . Selena's Work Drake's Work Who is correct and why? Selena is incorrect b

ecause she should have substituted the values for the variables first, and then simplifed. Selena is correct because she simplified correctly and then evaluated correctly after substituting the values for the variables. Drake is incorrect because he should have simplified first, before substituting the values for the variables. Drake is correct because he substituted the values for the variables first, and simplified correctly. Mark this and return
Mathematics
2 answers:
Anit [1.1K]3 years ago
8 0

Answer:

The answer is B

Step-by-step explanation:

:D

Shtirlitz [24]3 years ago
7 0

Answer:

Selena is correct because she simplified correctly and then evaluated correctly after substituting the values for the variables.

I did this lesson i asked my teacher to help me but i did this problem on my own so i hope its right :)

Step-by-step explanation:

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Step-by-step explanation:

Left hand side:

4 [sin⁶ θ + cos⁶ θ]

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4 [(sin² θ)³ + (cos² θ)³]

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4 [(sin² θ + cos² θ) (sin⁴ θ − sin² θ cos² θ + cos⁴ θ)]

Pythagorean identity:

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4 [sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ − 3 sin² θ cos² θ]

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A study was conducted and two types of engines, A and B, were compared. Fifty experiments were performed using engine A and 75 u
USPshnik [31]

Answer:

a) -6 mpg.

b) 2.77 mpg

c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Gas mileage A: Mean 36, standard deviation 6, sample of 50:

So

\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485

Gas mileage B: Mean 42, standard deviation 8, sample of 50:

So

\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314

Distribution of the difference:

Mean:

\mu = \mu_A - \mu_B = 36 - 42 = -6

Standard error:

s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142

A. Find the point estimate.

This is the difference of means, that is, -6 mpg.

B. Find the margin of error

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = zs = 1.96*1.4142 = 2.77

The margin of error is of 2.77 mpg

C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)

The lower end of the interval is the sample mean subtracted by M. So it is -6 - 2.77 = -8.77 mpg

The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg

The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

8 0
2 years ago
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