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nataly862011 [7]
3 years ago
9

7. Solve for x.V

t{?} " align="absmiddle" class="latex-formula">
x+ 3 = 20​
Mathematics
1 answer:
ch4aika [34]3 years ago
6 0
7+3=20

X+3=20
-3 -3
+3 & -3 cancel out
Then 20 minus 3 equals 7
So x=7
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Please answer thank you
ziro4ka [17]
Sorry thats a tough one
3 0
3 years ago
Which expression is the simplest form of (27x^-9)^1/3?
lana [24]
\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\sqrt[n]{a}=a^\frac{1}{n}\\\\(a^n)^m=a^{nm}\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(27x^{-9})^\frac{1}{3}=27^\frac{1}{3}(x^{-9})^\frac{1}{3}=\sqrt[3]{27}\cdot x^{-9\cdot\frac{1}{3}}=3x^{-3}=\dfrac{3}{x^3}
8 0
3 years ago
HELP PLEASE!!!!!
Furkat [3]

Answer:

b

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
(4.1.4) Let X and Y be Bernoulli random variables. Let Z = X + Y. a. Show that if X and Y cannot both be equal to 1, then Z is a
Fynjy0 [20]

Step-by-step explanation:

Given that,

a)

X ~ Bernoulli (p_x) and Y ~ Bernoulli (y_x)

X + Y = Z

The possible value for Z are Z = 0 when X = 0 and Y = 0

and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0

If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,

Therefore Z is a Bernoulli random variable

b)

If X and Y can not be both equal to  1

then,

p_z = P(X=1 or Y=1)\\

p_z = P(X=1)+P(Y=1)-P(=1 and Y =1)

p_z = P(x=1)+P(Y=1)\\\\p_z=p_x+p_y

c)

If both X = 1 and Y = 1 then Z = 2

The possible values of the random variable Z are 0, 1 and 2.

since a  Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution

7 0
2 years ago
Find the value of x and 3x.
kozerog [31]

Answer:

x = 90°

3x = 270°

Step-by-step explanation:

Split the shape in half through the plane of 3x and x.

We can find the angle of x/2 is 45°.

Since x/2 = 45°, then we can fid that x=90°.

To find 3x, we simply multiply 90° 3 times.

So 3x = 270°.

3 0
3 years ago
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