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Arisa [49]
3 years ago
14

The sum of 5 consecutive odd numbers is 145 what is the third number in this sequence ​

Mathematics
1 answer:
lawyer [7]3 years ago
8 0

9514 1404 393

Answer:

  29

Step-by-step explanation:

The third in sequence will be the middle one, the average value of all the numbers in the sequence. That average is the sum (145) divided by the number of numbers (5).

  third number = 145/5 = 29

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Which expression represents 75 divided by an unknown number? A. 75 – r B. 75 · r C. 75 ÷ r D. 75 + r
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Find f(-2) if f(x) = 4x + 8
zzz [600]

Answer:

0

Step-by-step explanation:

Use the hint lol
Step 1: Plug in -2 into the equation.
f(-2) = 4(-2) + 8


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2 years ago
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Find the midpoint of side EF. (4 points)
umka2103 [35]

Answer:

Midpoint of side EF would be (-.5,4.5)

Step-by-step explanation:

We know that the coordinates of a mid-point C(e,f) of a line segment AB with vertices A(a,b) and B(c,d) is given by:

e=a+c/2,f=b+d/2

Here we have to find the mid-point of side EF.

E(-2,3) i.e. (a,b)=(2,3)

and F(1,6) i.e. (c,d)=(1,6)

Hence, the coordinate of midpoint of EF is:

e=-2+1/2, f=3+6/2

e=-1/2, f=9/2

e=.5, f=4.5

SO, the mid-point would be (-0.5,4.5)

 

3 0
2 years ago
1. Review the lesson above and use the guided notes to help you understand.
Nezavi [6.7K]
I have completed the 1st one and have attached it.
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3 0
3 years ago
On a particular production line,the likelihood that a light bulb is defective is 5%. Ten light bulbs are randomly selected. What
KengaRu [80]

Answer:

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

Step-by-step explanation:

Consider the provided information,

Let X is the number of defective bulbs.

Ten light bulbs are randomly selected.

The likelihood that a light bulb is defective is 5%.

Therefore sample size is = n = 10

Probability of a defective bulb = p = 0.05.

Therefore, q = 1 - p = 1 - 0.05 = 0.95

Mean of binomial random variable: \mu=np

Therefore, \mu=10(0.05)=0.5

Variance of binomial random variable: \sigma^2=npq

Therefore, \sigma^2=10(0.05)(0.95)=0.475

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

5 0
3 years ago
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