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garik1379 [7]
3 years ago
6

Determine the value of the following:

Mathematics
1 answer:
Airida [17]3 years ago
7 0

Answer:

C=\left[\begin{array}{cc}1&4\\1&6\end{array}\right]

Step-by-step explanation:

<u>Matrix Addition</u>

Given two matrices A and B, the sum of A+B can only be defined if both matrices are of the same order, i.e. they have the same number of files and columns. If that was the case, then the sum of A and B, call it C is defined as

c_{i,j}=a_{i,j}+b_{i,j}

In other words, the resulting sum at row i and column j is the sum of the elements in that exact position

We are given the matrices

\left[\begin{array}{cc}-2&3\\2&4\end{array}\right] +\left[\begin{array}{cc}3&1\\-1&2\end{array}\right]

Adding them at the same position, we have

C=\left[\begin{array}{cc}-2+1&3+1\\2-1&4+2\end{array}\right]

C=\left[\begin{array}{cc}1&4\\1&6\end{array}\right]

Select option c.

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What is the excluded value of a/b-2
Readme [11.4K]

The correct excluded value is 2.

8 0
3 years ago
One year Ron had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched)
Arada [10]

Answer:

Ron's ERA has a z-score of -2.03.

Karla's ERA has a z-score of -1.86.

Due to the lower z-score, Ron had a better yean than Karla relative to their peers.

Step-by-step explanation:

Z-score:

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 question:

Since the lower the ERA, the better the pitcher, whoever's ERA has the lower z-score had the better year relative to their peers.

Ron

ERA of 3.06, so X = 3.06

For the males, the mean ERA was 5.086 and the standard deviation was 0.998. This means that \mu = 5.086, \sigma = 0.998

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.06 - 5.086}{0.998}

Z = -2.03

Ron's ERA has a z-score of -2.03.

Karla

ERA of 3.28, so X = 3.28

For  the females, the mean ERA was 4.316 and the standard deviation was 0.558. This means that \mu = 4.316, \sigma = 0.558

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.28 - 4.316}{0.558}

Z = -1.86

Karla's ERA has a z-score of -1.86.

Which player had the better year relative to their peers, Ron or Karla?

Due to the lower z-score, Ron had a better yean than Karla relative to their peers.

5 0
3 years ago
(6x2 - 2x) + (5x-7)
Black_prince [1.1K]

Answer:

D. 6x2 + 3x - 7

Step-by-step explanation:

7 0
2 years ago
The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha
Xelga [282]

Answer:

The average value of f over the interval [-2,5] is \frac{10}{7}.

Step-by-step explanation:

Let suppose that function f is continuous and integrable in the given intervals, by integral definition of average we have that:

\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6 (1)

\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20 (2)

By Fundamental Theorems of Calculus we expand both expressions:

\frac{F(3)-F(-2)}{3-(-2)} = -6

F(3) - F(-2) = -30 (1b)

\frac{F(5)-F(3)}{5-3} = 20

F(5) - F(3) = 40 (2b)

We obtain the average value of f over the interval [-2, 5] by algebraic handling:

F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)

F(5) - F(-2) = 10

\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}

\bar f = \frac{10}{7}

The average value of f over the interval [-2,5] is \frac{10}{7}.

4 0
2 years ago
Solve for u
Sergio039 [100]
U/9 = 8/12 u = 6

Step 1: Cancel the common factor (4)

u = 2
—- —-
9 3


Step 2: multiply both sides by 9


9u 2 * 9
—- = ——-
9 3

Step 3: simplify

2 *9 = 18

18 ÷ 3 = 6

u = 6
8 0
2 years ago
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