All categories

3 years ago

The month of the year is December. What is the type of data is December?

Mathematics

2 answers:

ExtremeBDS [4]3 years ago

6 0

December is categorical data. Hope this helps, if you have any questions please feel free to message me;)

scoundrel [369]3 years ago

3 0

Data that is collected can be either categorical or numerical data.Categorical variables represent types of data which may be divided into groups.

December is the twelfth and last month of the year .The word December is a categorical discrete data which is also known as qualitative, because it is a name instead of a number.Among the options given December is a month of a year.

The month of the year is December is a categorical data .

You might be interested in

What is the answer to 7m-6,m=8

nydimaria [60]

Answer:

Step-by-step explanation:

When given the following equation, and information,

7m - 6

m = 8

One must substitute the value of (m) into the equaiton. Then simplify and solve to evaluate the equation,

7m - 6

7(8) - 6

56 - 6

3 0

3 years ago

Read 2 more answers

Evaluate 0.3y+ y/z when y= 10 and z=5

vodka [1.7K]

Answer: answer is 5

Step-by-step explanation:see attachment

6 0

4 years ago

A group of 120 students were asked how old they were when they lost their first tooth. 50% said they were 7 years old. How many

Marat540 [252]

60 students, because 120 divided by 2 is 60, and 100% divided by 2 is 50%.

3 0

3 years ago

Read 2 more answers

Suppose that, for every lot of 100 computer chips a company produces, an average of 1.4 are defective. Another company buys many

Vesna [10]

Answer:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)$

And we can find the individual probabilities like this:

$P(X=0) =1.4^0 \frac{e^{-1.4}}{0!}= 0.2466$

$P(X=1) =1.4^1 \frac{e^{-1.4}}{1!}= 0.3452$

$P(X=2) =1.4^2 \frac{e^{-1.4}}{2!}= 0.2417$

$P(X=3) =1.4^3 \frac{e^{-1.4}}{3!}= 0.1128$

And replacing we got:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)=0.2466+ 0.3452+0.2417+0.1128=0.9463$

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

$P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}$

And the parameter $\lambda=1.4$ represent the average ocurrence rate per unit of time.

Solution to the problem

For this case the batch would be rejected if we found more than 3 defects, so then the probability of accept the batch would be given by:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)$

And we can find the individual probabilities like this:

$P(X=0) =1.4^0 \frac{e^{-1.4}}{0!}= 0.2466$

$P(X=1) =1.4^1 \frac{e^{-1.4}}{1!}= 0.3452$

$P(X=2) =1.4^2 \frac{e^{-1.4}}{2!}= 0.2417$

$P(X=3) =1.4^3 \frac{e^{-1.4}}{3!}= 0.1128$

And replacing we got:

$P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)=0.2466+ 0.3452+0.2417+0.1128=0.9463$

5 0

3 years ago

SUPER EASY AND ILL GIVE BRAINLIEST

Anna35 [415]

Answer:

You cannot put this in an input output chart because one value of x has 2 y values.

Step-by-step explanation:

8 0

3 years ago