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GarryVolchara [31]
3 years ago
9

3 + b/9 < 4show work please

Mathematics
1 answer:
Gekata [30.6K]3 years ago
8 0

First, before multiplying to cancel the fraction, let's isolate x by subtracting both sides by 3:

b/9 < 1

To cancel out the fraction, we multiply both sides by 9:

x < 9

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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
Kim got her pay raise, and her hourly pay went from $12 to $14. Jane also got a raise, and her hourly pay raise went from $13 to
Contact [7]
Kim because just cause man. 14/12 can be turned into 7/6... boom knowledge??
7 0
3 years ago
Read 2 more answers
Find the slope of the line.
ladessa [460]
Lets take any two points on the line . I took (-2,-2) and (0,-3)
slope is rise/run = -3-(-2)/ 0-(-2) = -3+2/2= -1/2
answer is -1/2
4 0
3 years ago
Joy has 15 coins with a value of $1.10 (nickels and dimes). how many of each are there?
Goryan [66]

Dimes= $0.10

Nickels=$0.05

$1.10 could hold 11 dimes or 22 nickels, but there on only 15 coins are in this total.

So a total of 7 dimes and 8 nickels is 15 coins and equal to $1.10


5 0
3 years ago
Read 2 more answers
A sample of Iron-59 contains 50 mg. One hundred days later, the sample contains 10.75 mg. What is the half-life of iron-59, in d
exis [7]

Answer:

t_{1/2}=45.09 d    

Step-by-step explanation:

We can use the decay equation:

M=M_{0}e^{-\lambda t}

  • M(0) is the initial mass
  • M is the mass after t (1000 days) time
  • λ is the decay constant

But:

\lambda = ln(2)/t_{1/2}

  • t(1/2) is the half-life.

So, we can rewrite the initial equation:

M=M_{0}e^{-\frac{ln(2)}{t_{1/2}}t}

Now, we just need to solve it for t(1/2):

ln(\frac{M}{M_{0}})=-\frac{ln(2)}{t_{1/2}}t

t_{1/2}=-\frac{ln(2)}{ln(\frac{M}{M_{0}})}t

t_{1/2}=-\frac{ln(2)}{ln(\frac{10.75}{50})}100                

t_{1/2}=45.09 d    

I hope it helps you!

5 0
2 years ago
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