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Musya8 [376]
2 years ago
14

Let g(x, y) = cos(x + 2y). (a) Evaluate g(2, -1). (b) Find the domain of g. (c) Find the range of g. 1. Find and sketch the doma

in of the function f(x, y) =sqrt(y-x^2)/(1-x^2) 2. Draw a contour map of the function f(x, y) = ye^-x showing several level curves. 3. Sketch both a contour map and a graph of the function f(x, y) = x^2 + 9y^2.
Mathematics
1 answer:
skelet666 [1.2K]2 years ago
4 0

Not going to do (1), (2), (3) since they are implicit from (a) and (b) and require graphs (which aren't supported on brainly).

We have g(x,y)=\cos(x+2y), that means, if plug in x=2,y=-1 we get g(2,-1)=\cos(2+2(-1))=\cos(0)=\boxed{1}.

Doman of cosine is x\in[0,1].

Hope this helps.

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What is 6,224 in scientific notation
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Answer:

6.224  x 10^3 Hope this helped out :)

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the  

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In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were re
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Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μemployed(μ1) be the sample mean of students who are employed and μnot employed(μ2) be the sample mean of students who are not employed

The random variable is μ1 - μ2 = difference in the mean of the employed and unemployed students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 < μ2 H1 : μ1 - μ2 < 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 3.22

μ2 = 3.33

s1 = 0.475

s2 = 0.524

n1 = 172

n2 = 116

t = (3.22 - 3.33)/√(0.475²/172 + 0.524²/116)

t = - 1.81

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.475²/172 + 0.524²/116]²/[(1/172 - 1)(0.475²/172)² + (1/116 - 1)(0.524²/116)²] = 0.00001353363/0.00000005878

df = 230

We would determine the probability value from the t test calculator. It becomes

p value = 0.036

Since alpha, 0.05 > than the p value, 0.036, then we would reject the null hypothesis.

Therefore, at a 5% significant level, this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed

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