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Mandarinka [93]
3 years ago
12

A company has actual unit demand for five consecutive years of 100, 105, 135, 145, and 150. The respective forecasts were 110 fo

r all five years. Which of the following is the resulting MAD value that can be computed from this data?
a. 23
b. 22
c. 24
d. 21
e. 30
Mathematics
1 answer:
atroni [7]3 years ago
6 0

Answer:

Option A) 23

Step-by-step explanation:

We are given the following data in the question:

100, 105, 135, 145, 150

The respective forecasts were 110 for all five years.

We have to calculate MAD value.

Deviations = 10, 5, 25, 35, 40

Sum of Deviations = 10+5+25+35+40 = 115

MAD =

\dfrac{\text{Sum of absolute deviation}}{5} = \dfrac{10+5+25+35+40}{5} = \dfrac{115}{5} = 23

Thus, the correct answer is

Option A) 23

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Which expression is equivalent to 3x + 5x( x - 2)
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I think it is c but pet me know

7 0
3 years ago
Given the function f(x) = 2^x, find the value of f^−1(32).
klio [65]
f(x)= 2^{x}

We are to find the value of f^{-1}(32)

First we need to find the inverse of f(x) . The inverse of 2^{x} is \frac{ln(x)}{ln(2)}.

Steps to find inverse of f(x) are shown below in the image.

So f^{-1}(32) = \frac{ln(32)}{ln(2)}=5

Therefore, the correct answer is option C

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3 years ago
Given that a*b = 2ab - 3b. Determine 2*(3*5)​
Oksanka [162]

Answer:

15

Problem:

Given that a*b = 2ab - 3b. Determine 2*(3*5)​

Step-by-step explanation:

First we find 3*5 using rule a*b = 2ab - 3b.

3*5=2(3)(5)-3(5)

3*5=30-15

3*5=15

Now we are left with 2*15 in which we will still use the rule a*b=2ab-3b.

2*15=2(2)(15)-3(15)

2*15=60-45

2*15=15

Therefore, 2*(3*5)=15.

3 0
3 years ago
Helppppppp<br> 3<br> Σ (2n – 3)<br> n=0
Nataly [62]

Since there are only 4 terms in the sum, it's not too much work to expand it as

\displaystyle \sum_{n=0}^3 (2n-3) = -3 + (-1) + 1 + 3 = \boxed{0}

Alternatively, we can use the well-known formulas

\displaystyle \sum_{n=1}^N 1 = \underbrace{1 + 1 + \cdots + 1}_{N \text{ 1s}} = N

\displaystyle \sum_{n=1}^N n = 1+2+\cdots+N = \frac{N(N+1)}2

These sums start at n = 0, so in our given sum we will keep track of the 0-th term separately:

\displaystyle \sum_{n=0}^3 (2n-3) = -3 + \sum_{n=1}^3 (2n-3) = -3 + 2 \sum_{n=1}^3 n - 3 \sum_{n=1}^3 1 = -3 + 3\times4 - 3\times3 = 0

as expected.

8 0
2 years ago
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