The matrix exists as a set of numbers placed in rows and columns to create a rectangular array. The manager could achieve scalar multiplication on Matrix A, utilizing the scalar 1.15.
<h3>What is the matrix?</h3>
The matrix exists as a set of numbers placed in rows and columns to create a rectangular array. The numbers exist named the elements, or entries, of the matrix. Matrices contain wide applications in engineering, physics, economics, and statistics as well as in different branches of mathematics.
Increasing the price by 15% would mean we exist taking 100% of the value + another 15%
100 + 15 = 115%
115% = 115/100 = 1.15.
Multiplying every value in Matrix A by 1.15 will give the price raised by 15%.
Therefore, the correct answer is option c. by multiplying each entry of matrix a by 1.15.
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The absolute value of debt of $45 is 45 because a balance of $0 on an account means there is no debt, and a negative of 45 value represents the debt.
<h3>What is debt?</h3>
It is defined as the amount one party needs to pay to another party as the first party borrowed an amount that will be credited by the second party. Debt occurs when one party cannot be able to purchase something under normal circumstances.
As we know,
The absolute value of a negative account balance is equal to the amount of debt. A balance of $0 on an account means there is no debt. Credit is shown by a positive account balance.
Richard has a debt of $45.
The negative of 45 value represents the debt.
-$45 = debt
Absolute value = |-45| = 45
Thus, the absolute value of debt of $45 is 45 because a balance of $0 on an account means there is no debt, and a negative of 45 value represents the debt.
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2t + 4 < -8
2t < -12
t < -6
For the inequality to be true, t must be a number that is less than -6.
Hope this helps!
Answer:
Mean: 40.17 years.
Standard deviation: 10.97 years.
Step-by-step explanation:
The frequency distribution is in the attached image.
We can calculate the mean adding the multiplication of midpoints of each class and frequency, and dividing by the sample size.
The midpoints of a class is calculated as the average of the bounds of the class.
Then, the mean can be written as:
The standard deviation can be calculated as:
![s=\sqrt{\dfrac{1}{N-1}\sum f_i(X_i-E(X))^2}\\\\\\s=\sqrt{\dfrac{1}{59}[3(15-40.17)^2+7(25-40.17)^2+18(35-40.17)^2+20(45-40.17)^2+12(55-40.17)^2]}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7BN-1%7D%5Csum%20f_i%28X_i-E%28X%29%29%5E2%7D%5C%5C%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7B59%7D%5B3%2815-40.17%29%5E2%2B7%2825-40.17%29%5E2%2B18%2835-40.17%29%5E2%2B20%2845-40.17%29%5E2%2B12%2855-40.17%29%5E2%5D%7D)
