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2 years ago

CAN SOMEONE PLEASE HELP ME WITH THIS ILL GIVE YOU BRAINLY IF YOU GET IT RIGHT ALSO PLEASE EXPLAIN HOW YOU GOT THE ANSWER!!!

Mathematics

1 answer:

sesenic [268]2 years ago

8 0

Answer:

C. P(E) = 0.9

Step-by-step explanation:

If E needs to be greater than D, then you have two options. 1.5 and 0.9. Since D is 0.5, I think that the most reasonable answer is 0.9.

[If this is wrong, the you know your answer is B. P(E) = 1.5]

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For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Anni [7]

Answer:

The reduced row-echelon form of the linear system is $\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

Interchange two rows
Multiply one row by a nonzero number
Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

$\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$

You need to follow these steps:

Divide row 1 by 2 $\left(R_1=\frac{R_1}{2}\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]$

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]$

Subtract row 1 multiplied by 5 from row 3 $\left(R_3=R_3-\left(5\right)R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 1 $\left(R_1=R_1-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 3 $\left(R_3=R_3-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]$

Multiply row 2 by 2 $\left(R_2=\left(2\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]$

Divide row 3 by −19 $\left(R_3=\frac{R_3}{-19}\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]$

Subtract row 3 multiplied by 16 from row 1 $\left(R_1=R_1-\left(16\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]$

Add row 3 multiplied by 6 to row 2 $\left(R_2=R_2+\left(6\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

8 0

3 years ago

What is 4/25 times 15/16

Mumz [18]

In a decimal it’s 0.15

7 0

2 years ago

Read 2 more answers

Helpppppp me with question 4 Thursday pleaseeeeeeeeeeeeeeee

Gennadij [26K]

If x is 0 then I think it would be the 3 to the second power and that is 9. Because 0 + 9 = 9. Correct me if I’m wrong!

8 0

2 years ago

I need help please?!!!!

sergejj [24]

A. i’m pretty sure !!

6 0

3 years ago

Read 2 more answers

The number of customers that visit a local small business is 51,200 and has been continuously declining at a rate of 3.8% each y

Hunter-Best [27]

Answer:

Final amount of customers =30141.44

Step-by-step explanation:

Amount of customer remaining

A= p(1-r/n)^(nt)

P= initial amount of customers

R= rate but it's a negative rate

N= number of times

T= number of years

A= final amount of customers

A= p(1-r/n)^(nt)

A= 51200(1-0.038/14)^(14*14)

A= 51200(1-0.0027)^196

A= 51200(0.9973)^196

A= 51200(0.5887)

A= 30141.44

6 0

2 years ago