How many times does to go into 50000

What is the solution to the following system?

leva [86]

Answer:

(4,3,2)

Step-by-step explanation:

We can solve this via matrices, so the equations given can be written in matrix form as:

$\left[\begin{array}{cccc}3&2&1&20\\1&-4&-1&-10\\2&1&2&15\end{array}\right]$

Now I will shift rows to make my pivot point (top left) a 1 and so:

$\left[\begin{array}{cccc}1&-4&-1&-10\\2&1&2&15\\3&2&1&20\end{array}\right]$

Next I will come up with algorithms that can cancel out numbers where R1 means row 1, R2 means row 2 and R3 means row three therefore,

-2R1+R2=R2 , -3R1+R3=R3

$\left[\begin{array}{cccc}1&-4&-1&-10\\0&9&4&35\\0&14&4&50\end{array}\right]$

$\frac{R_2}{9}=R_2$

$\left[\begin{array}{cccc}1&-4&-1&-10\\0&1&\frac{4}{9}&\frac{35}{9}\\0&14&4&50\end{array}\right]$

4R2+R1=R1 , -14R2+R3=R3

$\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&-\frac{20}{9}&-\frac{40}{9}\end{array}\right]$

$-\frac{9}{20}R_3=R_3$

$\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&1&2\end{array}\right]$

$-\frac{4}{9}R_3+R_2=R2$ , $-\frac{7}{9}R_3+R_1=R_1$

$\left[\begin{array}{cccc}1&0&0&4\\0&1&0&3\\0&0&1&2\end{array}\right]$

Therefore the solution to the system of equations are (x,y,z) = (4,3,2)

Note: If answer choices are given, plug them in and see if you get what is "equal to". Meaning plug in 4 for x, 3 for y and 2 for z in the first equation and you should get 20, second equation -10 and third 15.

7 0

3 years ago

An airline knows that 5 percent of the people making reservations on a certain flight will not show up. Consequently, their poli

kenny6666 [7]

Answer:

the probability that there will be a seat available for every passenger who shows up is 0.74

Step-by-step explanation:

if the probability of choosing a passenger that will not show up is 5% , then the probability of choosing a passenger that will show up is 95%.

Denoting event X= x passengers will show up from the total of 52 that purchased the ticket

Then P(X) follows a binomial probability distribution, since each passenger is independent from others. Thus

P(X) = n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= total number of passengers hat purchased the ticket= 52

p= probability of choosing a passenger that will show up

x = number of passengers that will show up

therefore the probability that the number does not exceed the limit of 50 passengers is:

P(X≤50)= ∑P(X=i) from i=0 to i=50 =F(50)

where F(x) is the cumulative binomial probability distribution, then from tables:

P(X≤50)= F(50) = 0.74

therefore the probability that there will be a seat available for every passenger who shows up is 0.74

8 0

3 years ago

Jessica and Nancy are members of different video game libraries. Jessica pays a membership fee of $40, and she pays $5 for every

DaniilM [7]

36 because I did this on my test before

5 0

3 years ago

How do you solve for X?<br> BD= 5x+3 AE= 4x+18

Kaylis [27]

Hope this helps :)))

6 0

3 years ago

A group of high school athletes has an average GPA of 2.7 with a standard deviation of 0.9. a)Find the percentage of athletes wh

mart [117]

Answer:

a) The percentage of athletes whose GPA more than 1.665 is 87.49%.

b) John's GPA is 3.645.

Step-by-step explanation:

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 2.7, \sigma = 0.9$