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zavuch27 [327]
3 years ago
15

What is the solution of - 8x- 5+3x=7+4x- 9

Mathematics
1 answer:
Ray Of Light [21]3 years ago
6 0
9x = -3 put together all your like terms
  -3/9 = x or -1/3 isolate x
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What are the terms in the expression 4 + 9V +6w?
spayn [35]

Answer:

4, 9v, and 6w

Step-by-step explanation:

I'm pretty sure it's right

4 0
2 years ago
How to rationalize the denominator?​
VMariaS [17]

Step-by-step explanation:

So, in order to rationalize the denominator, we need to get rid of all radicals that are in the denominator.

Step 1: Multiply numerator and denominator by a radical that will get rid of the radical in the denominator. ...

Step 2: Make sure all radicals are simplified. ...

Step 3: Simplify the fraction if needed.

3 0
3 years ago
Which fraction is closest to 1/2? A number line from negative 1 to 1 in halves. 1/6 3/8 3/4 ‐1/2
Arturiano [62]
Answer: 3/8

because 3/8 is .375
And 3/4 is .75
And 1/6 is .1666

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5 0
2 years ago
JK has endpoints at J(5,9) and K(7, 7). Find the midpoint M of JK.
Basile [38]

Answer:

J (7, 10)

K (3, 2)

Formula for midpoint:

x - coordinate for M:

y - coordinate for M:

Therefore, M (5,6).

Hope this helps :)

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
(a) A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure
Temka [501]

Answer:

For first lamp ; The resultant probability is 0.703

For both lamps; The resultant probability is 0.3614

Step-by-step explanation:

Let X be the lifetime hours of two bulbs

X∼exp(1/1400)

f(x)=1/1400e−1/1400x

P(X<x)=1−e−1/1400x

X∼exp⁡(1/1400)

f(x)=1/1400 e−1/1400x

P(X<x)=1−e−1/1400x

The probability that both of the lamp bulbs fail within 1700 hours is calculated below,

P(X≤1700)=1−e−1/1400×1700

=1−e−1.21=0.703

The resultant probability is 0.703

Let Y be a lifetime of another lamp two bulbs

Then the Z = X + Y will follow gamma distribution that is,

X+Y=Z∼gamma(2,1/1400)

2λZ∼

X+Y=Z∼gamma(2,1/1400)

2λZ∼χ2α2

The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,

P(Z≤1700)=P(1/700Z≤1.67)=

P(χ24≤1.67)=0.3614

The resultant probability is 0.3614

8 0
2 years ago
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