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wolverine [178]
3 years ago
13

Can someone please help me answer this Math problem?

Mathematics
1 answer:
kap26 [50]3 years ago
4 0

Answer:

2√2.

Step-by-step explanation:

We use the Pythagoras Theorem:

x^2 = 2^2 + 2^2

x^2 =  8

x = √8

= 2√2.

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Suppose a polling agency reported that 45.7% of registered voters were in favor of raising income taxes to pay down the national
sineoko [7]
<h2>Answer with explanation:</h2>

As per given , we have

p= 0.457

n= 1033

Margin of error for 99​% confidence : E= 4%=0.04

Confidence interval : p\pm E

i.e. 0.457\pm 0.04

=( 0.457- 0.04,\ 0.457+0.04)=(0.417,\ 0.497)

99% confidence interval for the proportion of registered voters who are in favor of raising income taxes to pay down the national debt. = (0.417, 0.497)

Interpretation : We are 99% sure that the true population proportion of registered voters who are in favor of raising income taxes to pay down the national debt lies in interval (0.417, 0.497).

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3 years ago
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The answer is 2.6. I could be wrong.
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Sliva [168]

A differentiable function f(x) is increasing on an open interval (a,b) if f'(x)>0 for all a, and decreasing if f'(x). For this problem, you then need to compute the derivative:

f(x)=x^2\ln x\implies f'(x)=2x\ln x+x=(2\ln x+1)x

then solve for f'(x)=0:

(2\ln x+1)x=0\implies x=0\text{ or }x=e^{-1/2}

We can ignore x=0 because x^2\ln x is defined only for x>0. So we have two intervals to consider, (0,e^{-1/2}) and (e^{-1/2},\infty). All we need to do is pick any value from either interval and check the sign of the derivative f'(x). Since e^{-1/2}\approx0.606, from the first interval we can take x=\dfrac12, and from the second we can pick x=1.

f'\left(\dfrac12\right)\approx-0.193

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The above indicates that f(x) is decreasing on the first interval (0,e^{-1/2}), and increasing on the second interval (e^{-1/2},\infty).

For part (b), we use the info from above as part of the first derivative test for extrema. We have one critical point at x=e^{-1/2}, and we know how f(x) behaves to either side of this point; f(x) decreases to left of it, and increases to the right. This pattern is indicative of a minimum occurring at x=e^{-1/2}, and we find that f(x) has the (local) minimum value of f(e^{-1/2})=-\dfrac1{2e}\approx-0.184.

8 0
3 years ago
How do you find the angle measure of a regular dodecagon?
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