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3 years ago

Is the function even, odd, or neither?

Mathematics

2 answers:

Step2247 [10]3 years ago

3 0

Answer:

A function with a graph that is symmetric about the origin is called an odd function. Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, f(x)=2x f ( x ) = 2 x is neither even nor odd.

Step-by-step explanation:

enot [183]3 years ago

3 0

You may be asked to "determine algebraically" whether a function is even or odd. To do this, you take the function and plug –x in for x, and then simplify. If you end up with the exact same function that you started with (that is, if f (–x) = f (x), so all of the signs are the same), then the function is even. If you end up with the exact opposite of what you started with (that is, if f (–x) = –f (x), so all of the signs are switched), then the function is odd.

In all other cases, the function is "neither even nor odd".

Let's see what this looks like in action:

Determine algebraically whether f (x) = –3x2 + 4 is even, odd, or neither.

If I graph this, I will see that this is "symmetric about the y-axis"; in other words, whatever the graph is doing on one side of the y-axis is mirrored on the other side:

graph of y = –3x^2 + 4

This mirroring about the y-axis is a hallmark of even functions.

Also, I note that the exponents on all of the terms are even — the exponent on the constant term being zero: 4x0 = 4 × 1 = 4. These are helpful clues that strongly suggest to me that I've got an even function here.

But the question asks me to make the determination algebraically, which means that I need to do the algebra.

So I'll plug –x in for x, and simplify:

f (–x) = –3(–x)2 + 4

= –3(x2) + 4

= –3x2 + 4

I can see, by comparing the original function with my final result above, that I've got a match, which means that:

f (x) is even

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Suppose you have the following information. 12% of all drivers do not have a valid driver’s license, 6% of all drivers have no i

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e) 0.14

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a driver does not have a valid driver's license.

B is the probability that a driver does not have insurance.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a driver does not have a valid driver's license but has insurance and $A \cap B$ is the probability that a driver does not have any of these things.

By the same logic, we have that:

$B = b + (A \cap B)$

We start finding these values from the intersection.

4% have neither

This means that $(A \cap B) = 0.04$

6% of all drivers have no insurance

This means that $B = 0.06$ . So

$B = b + (A \cap B)$

$0.06 = b + 0.04$

$b = 0.02$

12% of all drivers do not have a valid driver’s license

This means that $A = 0.12$

$A = a + (A \cap B)$

$0.12 = a + 0.04$

$a = 0.08$

The probability that a randomly selected driver either fails to have a valid license or fails to have insurance is about

$P = a + b + (A \cap B) = 0.08 + 0.02 + 0.04 = 0.14$

So the correct answer is:

e) 0.14

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3 years ago

Do teachers find their work rewarding and satisfying? An article reports the results of a survey of 397 elementary school teache

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Answer:

The 95% confidence interval would be given (0.0109;0.1651).

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for elementary school

$\hat p_A =\frac{226}{397}=0.569$ represent the estimated proportion for elementary school

$n_A=397$ is the sample size required for Brand A

$p_B$ represent the real population proportion for high school teachers

$\hat p_B =\frac{129}{268}=0.481$ represent the estimated proportion for high school teachers

$n_B=268$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.569-0.481) - 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.0109$

$(0.569-0.481) + 1.96 \sqrt{\frac{0.569(1-0.569)}{397} +\frac{0.481(1-0.481)}{268}}=0.1651$

And the 95% confidence interval would be given (0.0109;0.1651).

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$

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