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2 years ago

6

To the model estimated in table 8.1, add the interaction term, e401k · inc. estimate the equation by ols and obtain the usual an

d robust standard errors. what do you conclude about the statistical significance of the interaction term?

1 answer:

taurus [48]2 years ago

3 0

Please Send Lions, Monkeys, Cats And Zebras Into Lovely Hot Countries
Signed General Penguin
P - potassium
S - sodium
L - lithium
M - magnesium
C - calcium
A - aluminium
Z - zinc
I - iron
L - lead
H - hydrogen
C - copper
S - silver
G - gold
<span>P - platinum</span>

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Question 3 (1 point)

Virty [35]

Answer:

The answer to your question is (4, 6)

Step-by-step explanation:

Data

E ( 9 , 7 )

F ( - 1, 5)

Formula

$Xm = \frac{x1 + x2}{2}$

$Ym = \frac{y1 + y2}{2}$

Substitution and simplification

$Xm = \frac{9 -1}{2}$

$Xm = \frac{8}{2}$

Xm = 4

$Ym = \frac{7 + 5}{2}$

$Ym = \frac{12}{2}$

Ym = 6

Result

(4 , 6)

7 0

3 years ago

I will give BRAINLIEST! ANSWER FOR 30 Points !!!!

Lilit [14]

Answer:

The answer is 3

Step-by-step explanation:

You can see that it recurs faster than usual for a tan curve so 1 is eliminated and it cant be 2 or 4 because it isnt recurring equally.

<h2>Please give brainliest</h2>

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1 year ago

If a smoothie recipe calls for 8 pints of mango juice but only 10 cups of mango juice are available, what is the scaling factor

mezya [45]

The amount of recipes is 6

8 0

3 years ago

Read 2 more answers

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

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2 years ago

Read 2 more answers

Alex purchased a new car for $28000. The car depreciates 7.25% each year. what will the price be in 5 years?

dalvyx [7]

Use the depreciation formula.

$\sf p(1-r)^n$

Where 'p' is the principal value, 'r' is the rate it depreciates, and 'n' is the time. Just plug in what we know:

$\sf 28000(1-0.0725)^5$

Simplify by subtracting:

$\sf 28000(0.9275)^5$

Simplify the exponent:

$\sf 28000(0.6864)$

Multiply:

$\sf\approx\boxed{\sf\$ 19219.20}$

8 0

2 years ago