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just olya [345]
4 years ago
14

3x - 15 = 7 - 8x how do you solve this​

Mathematics
2 answers:
Sphinxa [80]4 years ago
8 0

Answer:

324

Step-by-step explanation:

3x3 = 9

9-15 = -6

8x8=64

7-64=-54

-54 x -6 = 324

Nikitich [7]4 years ago
6 0

Answer:

<em>X=2</em>

Step-by-step explanation:

3x-15=7-8x

3x+8x=7+15

11x=22

x=\frac{22}{11}

x=2

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3 years ago
Can someone explain how to do this? thank you
ella [17]

Answer:

В (50°)

Step-by-step explanation:

m∠XWY=180-95=85°

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7 0
3 years ago
Did tosh get his donut
sergeinik [125]
Yes he did, the proof is in the context
3 0
3 years ago
What is the simplified form of 3 Start Root 5 c End Root times Start Root 15 c cubed End root? (worth 30 brainily points)
aleksley [76]

Answer:

A. 15 c squared Start Root 3 End Root

Step-by-step explanation:

3\sqrt{5c} *\sqrt{15c^3} \\

Multiply the terms inside the square roots together

3\sqrt{75c^4}

Look for items that are perfects squares

3\sqrt{25*3*c^4}

We can separate terms inside the square root into separate square roots

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8 0
4 years ago
Suppose that the lifetime of tv tubes are normally distributed with a standard deviation of 1.1 years. Suppose that exactly 20%
IrinaK [193]

Answer:

4.924 years

Step-by-step explanation:

Lets denote X the lifetime of a tv tube (In years). X has distribution N(\mu, 1.1) , with \mu unknown.

We know that P(X < 4) = 0.2. Using this data, we can find the value of \mu throught standarization.

Lets call Z = \frac{X - \mu}{1.1} the standarization of X. Z has distribution N(0,1), and its cummulative function, \phi is tabulated. The values of \phi can be found in the attached file.

P(X < 4) = P(\frac{X - \mu}{1.1} < \frac{4 - \mu}{1.1}) = P(Z < \frac{4 - \mu}{1.1}) = \phi(\frac{4 - \mu}{1.1}) = 0.2

The value q such that \phi (q) = 0.2 doesnt appear on the table. We can find it by using the symmetry of the normal density function. The opposite of q, -q must verify that \phi(-q) = 0.8 , hence -q must be equal to 0.84. Thus, q = -0.84

But this value of q should match with the number \frac{4 - \mu}{1.1} , so we have

\frac{4- \mu}{1.1} = -0.84

4 - \mu = 1.1 * (-0.84) = -0.924

\mu = 4 - (-0.924) = 4.924

Thus, the expected lifetime of TV tubes is 4.924 years.

I hope this works for you!

Download pdf
7 0
4 years ago
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