I think (7,10)
I hope that I helped ;D
Answer:
r = 4.5%
Step-by-step explanation:
Formula:
I = Prt
r = I/(Pt)
Given:
I = 662.29
P = 4205
t = 3.5
Work:
r = I/(Pt)
r = 662.29/(4205 * 3.5)
r = 662.29/14717.5
r = 0.045
r = 4.5%
Answer:
The top 20% of the students will score at least 2.1 points above the mean.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

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.
The mean of a certain test is 14 and the standard deviation is 2.5.
This means that 
The top 20% of the students will score how many points above the mean
Their score is the 100 - 20 = 80th percentile, which is X when Z has a pvalue of 0.8. So X when Z = 0.84.
Their score is:




16.1 - 14 = 2.1
The top 20% of the students will score at least 2.1 points above the mean.
Answer:
The answer is "
".
Step-by-step explanation:
![\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D)
Solve the L.H.S part:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D)
After calculating the L.H.S part compare the value with R.H.S:
![\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D%20%5C%5C%5C%5C)

In equation (i) multiply by 3 and subtract by equation (iii):

put the value of c in equation (i):

In equation (ii) multiply by 3 then subtract by equation (iv):

put the value of d in equation (iv):

The final answer is "
".
<em>hey</em><em> </em><em>buddy</em><em>,</em>
<em>●</em><em>15x</em><em> </em><em>+</em><em> </em><em>5y</em><em> </em><em>=</em><em> </em><em>25</em>
<em>5y</em><em> </em><em>=</em><em> </em><em>25-15x</em>
<em>y</em><em> </em><em>=</em><em> </em><em>(</em><em>25</em><em> </em><em>-</em><em> </em><em>15x</em><em>)</em><em> </em><em>/</em><em>5</em>
<em>y</em><em> </em><em>=</em><em> </em><em>5</em><em> </em><em>-</em><em> </em><em>3x</em>