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joja [24]
3 years ago
14

How do you know the correct order in which to evaluate algebraic expressions?

Mathematics
2 answers:
aliya0001 [1]3 years ago
8 0
Put them in chronological order.
shtirl [24]3 years ago
4 0
You just see if they have the same variable or not.
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Is 1/2 greater than 2/8
kifflom [539]
Yes, because 2/8 simplified is 1/4.

1/2 > 1/4
8 0
3 years ago
Read 2 more answers
L has $200. Ty has 30% more than Lu and twice as much as Ali. How much money do they have altogether? (PLEASE SHOW STEPS!)
saul85 [17]
30% of 200=60 200+60= 260 
Ty has £260
£260 x 2= £520

200+260+520= £980

Answer £980

I hope this is right :D
5 0
3 years ago
What is the approximate circumference of a circle that has a radius of 6 units? Round your answer to the nearest tenth.
Angelina_Jolie [31]

Answer:

37.7 units²

Step-by-step explanation:

the formula for circumference of a circle is πr²

π x 6² = 37.69 ≈37.7²

8 0
2 years ago
4. A random variable X has a mean of 10 and a standard deviation of 3. If 2 is added to each value of X, what will the new mean
Ede4ka [16]

Adding 2 to each value of the random variable X makes a new random variable X+2. Its mean would be

E[X+2]=E[X]+E[2]=E[X]+2

since expectation is linear, and the expected value of a constant is that constant. E[X] is the mean of X, so the new mean would be

E[X+2]=10+2=12

The variance of a random variable X is

V[X]=E[X^2]-E[X]^2

so the variance of X+2 would be

V[X+2]=E[(X+2)^2]-E[X+2]^2

We already know E[X+2]=12, so simplifying above, we get

V[X+2]=E[X^2+4X+4]-12^2

V[X+2]=E[X^2]+4E[X]+4-12^2

V[X+2]=(V[X]+E[X]^2)+4E[X]-140

Standard deviation is the square root of variance, so V[X]=3^2=9.

\implies V[X+2]=(9+10^2)+4(10)-140=9

so the standard deviation remains unchanged at 3.

NB: More generally, the variance of aX+b for a,b\in\mathbb R is

V[aX+b]=a^2V[X]+b^2V[1]

but the variance of a constant is 0. In this case, a=1, so we're left with V[X+2]=V[X], as expected.

5 0
3 years ago
21 less than tree-fifths of a number
Oxana [17]

(3/5)x - 21

hope this helps :)

6 0
2 years ago
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