All categories

2 years ago

I'm having a bit of trouble finding out how to find the Domain of this function. Help please!

Mathematics

1 answer:

BabaBlast [244]2 years ago

6 0

In general, the domain is the set of all x-values for the graph.

The issue here is that this isn't the graph of a function. A function has at most one y-value for each x-value and this graph has an infinite number of y-values for the single x-value of 1.

So, either your teacher is wanting you to say the domain is {1}, because that's the only x-value used by the graph, or they're wanting you to say this is a trick question, because this isn't the graph of a function.

The range is the set of all y-values, which is -9<y<9, but again, do they intend this to be a trick question?

You might be interested in

The inverse of 2x^3+3

Rashid [163]

2/x ^6-3 you do the opposite of everything

6 0

3 years ago

Read 2 more answers

What best describes how to evaluate a variable expression?

Kaylis [27]

To replace each letter with its value, and then finish it by doing the order of the operation/ solve it.

7 0

3 years ago

What type of decimal is this fraction: LaTeX: \frac{11}{20}

skad [1K]

11/20

Divide 11 by 20 to find the decimal value.

11/20 = 0.55

The decimal only contains two numbers, so it is a terminating fraction.

6 0

3 years ago

An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi

jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable X denotes the number of heads in n tosses of an honest coin (n ≥ 30), then X has an approximately normal distribution with mean, $\mu=\frac{n}{2}$ and standard deviation, $\sigma=\frac{\sqrt{n}}{2}$ .

Here the number of tosses is, n = 2500.

Since n is too large, i.e. n = 2500 > 30, the random variable X follows a normal distribution.

The mean and standard deviation are:

$\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25$

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, μ = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

= μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

= 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

= 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

7 0

3 years ago

Use the cumulative normal distribution table to find the z-scores thqt bound the middle of 68% of theje area under the stanrard

padilas [110]

Answer:

Z scores between -0.995 and 0.995 bound the middle 68% of the area under the stanrard normal curve

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Middle 68%

Between the 50 - (68/2) = 16th percentile and the 50 + (68/2) = 84th percentile.

16th percentile:

X when Z has a pvalue of 0.16. So X when Z = -0.995

84th percentile:

X when Z has a pvalue of 0.84. So X when Z = 0.995.

Z scores between -0.995 and 0.995 bound the middle 68% of the area under the stanrard normal curve

7 0

3 years ago