All categories

3 years ago

Classify the random variables below according to whether they are discrete or continuous.

Mathematics

1 answer:

Ann [662]3 years ago

3 0

Answer:

a) Continuous

b) Continuous

c) Discrete

d) Continuous

e) Discrete

Step-by-step explanation:

Discrete variable

Discrete variables are numerical variables that have a countable number of values between any two values. A discrete variable is always numerical. For example, the number of people with blood type Upper A in a random sample of 42 people.

Continuous variable

Continuous variables are numerical variables that have an infinite number of values between any two values. A continuous variable can be numeric or date / time. For example, the time it takes to fly from City Upper A to City Upper B.

You might be interested in

Solve 6-7+3(-11+5)(-11+5)-5

andrew11 [14]

I think it's 112 I did it in my head might be wrong

6 0

3 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

8 0

3 years ago

Which is equivalent to 64 to the 1/4 power

Sladkaya [172]

Answer:

The answer to your question is the first option

Step-by-step explanation:

64 $^{1/4}$

Process

1.- Find the prime factors of 64

64 2

32 2

16 2

8 2

4 2

2 2

64 = 2⁶

2.- Express 64 as a fractional exponent

64 $^{1/4} = 2^{6/4}$

3.- Simplify

64 $^{1/4} = 2^{4/4} 2^{2/4}$

64 $^{1/4} = 2^{1} 2^{2/4}$

64 $^{1/4} = 2\sqrt[4]{2^{2}}$

4.- Result

64 $^{1/4} = 2\sqrt[4]{4}$

5 0

3 years ago

Two angles of a quadrilateral measure 150 degrees and 105 degrees the other two angles are in a ratio of 3:4 what are the measur

iragen [17]

Answer:

First angle is 45 and the second angle is 60

Step-by-step explanation:

150 + 105 = 255

360 - 255 = 105

the ratio is 3:4 I add the ratio so 7

105 / 7 = 15

The first angle with the ratio of 3 is 15 x 3 = 45

The second angle with the ratio of 4 is 4 x 15 = 60

7 0

3 years ago

5) dilation of 2 about the origin

Misha Larkins [42]

Answer:

the answer is c I did it before but I can't explain step by step I hope it helps u

5 0

3 years ago