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nignag [31]
3 years ago
6

PLEASE ANSWER RIGHT AWAY! What the LCM of 2, & 5

Mathematics
2 answers:
Mumz [18]3 years ago
8 0
10 is the lcm of those to numbers
kaheart [24]3 years ago
5 0
2,4,6,8,10
5,10

So ten is the lcm.
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Find the equation of a line through (2,-4) and (-7,-4)
steposvetlana [31]
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Find slope : 
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\text {Slope = }  \dfrac{-4 - (-4)}{-7-2}  =  \dfrac{0}{-9}  = 0

When the slope is 0, the graph is a horizontal line.

Equation of the graph is y = -4

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Answer: y = -4
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3 0
3 years ago
2(x+3) + 6 = 3x - 9 written in word form? HELP!!!
zheka24 [161]
Two multiplied by x plus three plus six is equal to three multiplied by x subtracted by nine
7 0
3 years ago
You order from a catalog the following office supplies: 10 boxes of paper, each box is 6 pounds and 36 boxes of pens, each box w
babymother [125]

$40.50 10 boxes at 6lbs = 60lbs

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6 0
3 years ago
if f(1) = -5 and f(n) = f(n-1) +7, find the first four terms and the common difference of the sequence
Alex_Xolod [135]

Answer:

- 5, 2, 9, 16 and d = + 7

Step-by-step explanation:

to obtain the first four terms substitute n = 2, 3, 4 into the recursive formula

f(1) = - 5 ← given

f(2) = f(1) + 7 = - 5 + 7 = 2

f(3) = f(2) + 7 = 2 + 7 = 9

f(4) = f(3) + 7 = 9 + 7 = 16

common difference d = 16 - 9 = 9 - 2 = 2 - (- 5) = 7



6 0
3 years ago
The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second. Use the central limi
Mila [183]

Answer:

0.048 is the probability that more than 950 message arrive in one minute.

Step-by-step explanation:

We are given the following information in the question:

The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second.

Let X be the number of messages arriving at a multiplexer.

Mean = 15

For poison distribution,

Mean = Variance = 15

\text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{15} = 3.872

From central limit theorem, we have:

z = \displaystyle\frac{x-n\mu}{\sigma\sqrt{n}}

where n is the sample size.

Here, n = 1 minute = 60 seconds

P(x > 950)

P( x > 950) = P( z > \displaystyle\frac{950 - (60)(15)}{\sqrt{(15)(60)}}) = P(z > 1.667)

= 1 - P(z \leq 1.667)

Calculation the value from standard normal z table, we have,  

P(x > 950) = 1 - 0.952 = 0.048

0.048 is the probability that more than 950 message arrive in one minute.

3 0
3 years ago
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