All categories

3 years ago

6. What are the dimensions of this matrix?

Mathematics

1 answer:

Svetlanka [38]3 years ago

5 0

Answer:

the dimensions will be 2 x 1

Option C is correct option.

Step-by-step explanation:

We need to determine the dimensions of this matrix $\left[\begin{array}{c}2&\sqrt{5} \end{array}\right]$

The dimensions consists of rows and columns in the form. Rows x Columns

In the given matrix the rows are: 2

The columns are: 1

So, the dimensions will be 2 x 1

Option C is correct option.

You might be interested in

Simplify the expression. Type your answer in the blank.<br> -

den301095 [7]

The abolute value of -13 is 13.
The negative value of 13 is -13.
Final answer: -13

3 0

4 years ago

Read 2 more answers

What is the product?

Alisiya [41]

Answer:

$\frac{1}{8}*(-\frac{3}{2})\\ =-\frac{3}{16}$

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Assume that power surges occur as a Poisson process with rate 3 per hour. These events cause damage to a computer, thus a specia

lidiya [134]

Answer:

a) 0.30327

b) 0.07582

Step-by-step explanation:

This is a Poisson distribution problem

Poisson distribution probability function is given as

P(X = x) = (e^-λ)(λˣ)/x!

where λ = mean = 3 power surges per hour

x = variable whose probability is required

a) If a single disturbance in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of a single disturbance (power surge) in 10 minutes?

λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.

P(X = x) = (e^-λ)(λˣ)/x!

λ = 0.5 disturbance per 10 minutes

x = 1 disturbance per 10 minutes

P(X = 1) = (e^-0.5)(0.5¹)/1! = (e⁻⁰•⁵)(0.5)/1!

= 0.30327

b) If two disturbances in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of two disturbances (power surges) in 10 minutes?

λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.

P(X = x) = (e^-λ)(λˣ)/x!

λ = 0.5 disturbance per 10 minutes

x = 2 disturbance per 10 minutes

P(X = 2) = (e^-0.5)(0.5²)/2! = (e⁻⁰•⁵)(0.25)/2!

= 0.07582

Hope this Helps!!!

3 0

4 years ago

50 POINTS !!<br><br><br> PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.

valentina_108 [34]

Answer: