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s344n2d4d5 [400]
4 years ago
14

The output of a process is stable and normally distributed. If the process mean equals 23.5, the percentage of output expected t

o be less than or equal to the mean: a. is 50%. b. is greater than 75%. c. cannot be determined without knowing the standard deviation value. d. is less than 25%
Mathematics
1 answer:
erastovalidia [21]4 years ago
6 0

Answer:

Option a) 50% of output expected to be less than or equal to the mean.

Step-by-step explanation:

We are given the following in the question:

The output of a process is stable and normally distributed.

Mean = 23.5

We have to find the percentage of output expected to be less than or equal to the mean.

Mean of a normal distribution.

  • The mean of normal distribution divides the data into exactly two equal parts.
  • 50% of data lies to the right of the mean.
  • 50% of data lies to the right of the mean

Thus, by property of normal distribution 50% of output expected to be less than or equal to the mean.

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4 0
3 years ago
Read 2 more answers
What is 6-2p=p+9 ?Because I don't know it​
nikklg [1K]

Hey there!

  • Firstly, you have to \bold{simplify} each of the sides of your equation
  • \bold{6-2p-p+9\downarrow}
  • \bold{6+(-2p)=p+9\downarrow}
  • \rightarrow\bold{-2p+6=p+9}
  • Now that you have that  portion out of the way, we could  move on to our second step... which is to \bold{subtract} by the value of  \bold{p} on each of your sides
  • \bold{-2p+6-p-p+9-p\downarrow}
  • \rightarrow\bold{-3p+6-9} (you get this because we solve the particular equation we added above)
  • Thirdly, you have to \bold{subtract} by the number \bold{6} on each of your sides
  • \bold{-3p+6-6=9-6\downarrow}
  • \bold{Cancel:6-6} because it equals to \bold{0}
  • \bold{Keep:9-6} because it brings us closer and closer to the answer of \bold{p}
  • Fourthly, we have to \bold{divide} by  the number \bold{-3} on each of your sides
  • \bold{\frac{-3p}{-3}=\frac{3}{-3}}
  • \boxed{\boxed{\bold{Answer:p=-1}}} \checkmark

Good luck on your assignment and enjoy your day!

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8 0
4 years ago
Consider the function f(x)=−3x+3 on the interval [−8,4]. Find the absolute extrema for the function on the given interval. Expre
Nikitich [7]

By means of <em>functions</em> theory and the characteristics of <em>linear</em> equations, the <em>absolute</em> extrema of the <em>linear</em> equation f(x) = - 3 · x + 3 are 27 (<em>absolute</em> maximum) for x = - 8 and - 9 (<em>absolute</em> minimum) for x = 4. (- 8, 27) and (4, - 9).

<h3>What are the absolute extrema of a linear equation within a closed interval?</h3>

According to the functions theory, <em>linear</em> equations have no absolute extrema for all <em>real</em> numbers, but things are different for any <em>closed</em> interval as <em>absolute</em> extrema are the ends of <em>linear</em> function. Now we proceed to evaluate the function at each point:

Absolute maximum

f(- 8) = - 3 · (- 8) + 3

f(- 8) = 27

Absolute minimum

f(4) = - 3 · 4 + 3

f(4) = - 9

By means of <em>functions</em> theory and the characteristics of <em>linear</em> equations, the <em>absolute</em> extrema of the <em>linear</em> equation f(x) = - 3 · x + 3 are 27 (<em>absolute</em> maximum) for x = - 8 and - 9 (<em>absolute</em> minimum) for x = 4. (- 8, 27) and (4, - 9).

To learn more on absolute extrema: brainly.com/question/2272467

#SPJ1

6 0
1 year ago
If k stands for an integer, then is it possible for k2 + k to stand for an odd integer? Be prepared to justify your answer.
BartSMP [9]

Answer:

<em>k^2 + k never stands for an odd integer</em>

Step-by-step explanation:

Let us consider either case with which k stands for an odd or even integer;

Case 1: k is an odd integer

For integer a, k = 2a + 1

So, k + 1 = 2a + 2 = 2( a + 1 ) = 2b for integer b

k^2 + k = k ( k + 1 ) = k ( 2b ) = 2kb = 2c for integer c,

<em>Therefore, if k is an odd integer, then k^2 + k is an even integer ;</em>

Case 2: k is an even integer

For an integer a, k = 2a

So, k + 1 = 2a + 1

k^2 + k = k( k+1 ) = 2a( 2a + 1 ) , multiple of 2

<em>Therefore, if k is an even integer, then k^2 + k is an even integer;</em>

<em>This would make k^2 + k never stand for an odd integer</em>

5 0
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