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MatroZZZ [7]
3 years ago
5

Round 854 to the nearest hundredth

Mathematics
1 answer:
Paladinen [302]3 years ago
4 0

Answer:

854

Step-by-step explanation:

There is no decimals.

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X^2-7x-34=10<br><br> Please help me answer this question thank you
elena-s [515]

Answer: x = 4,11

Step-by-step explanation:

x^2 - 7x - 34 - 10 = 0

x^2 - 7x - 44 = 0

Factorise it

x^2 - 11x - 4x - 44 = 0

x ( x - 11 ) - 4 ( x - 11 ) = 0

(x-4)(x-11) = 0

x = 4,11

6 0
1 year ago
Read 2 more answers
Help help please please
gtnhenbr [62]

✽ Hello there! ✽

4x+9<21  

Move all the constants (numbers) to the right, using the opposite                         operation:

4x<21-9

4x<12

Divide both sides by 4 to isolate x:

x<3

<h2>Therefore, x<3 ✔︎</h2>

Hope this helps!

~Just a felicitous girlie

#HaveASplendidDay

SilentNature

8 0
2 years ago
19) *
RUDIKE [14]
The correct answer is 18 sides.
7 0
3 years ago
Solve the linear equation 5(8x+2) -64=2(8x9) Find X
raketka [301]
Bla bla bla your answer is x=4.95

3 0
3 years ago
An individual who has automobile insurance from a certain company is randomly selected. Let y be the number of moving violations
Hoochie [10]

Answer:

a) E(Y)= \sum_{i=1}^n Y_i P(Y_i)

And replacing we got:    

E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95

b) E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".  

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).  

And the standard deviation of a random variable X is just the square root of the variance.  

Solution to the problem

Part a

We have the following distribution function:

Y        0         1         2       3

P(Y)  0.45    0.2    0.3   0.05

And we can calculate the expected value with the following formula:

E(Y)= \sum_{i=1}^n Y_i P(Y_i)

And replacing we got:    

E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95

Part b

For this case the new expected value would be given by:

E(80Y^2)= \sum_{i=1}^n 80Y^2_i P(Y_i)

And replacing we got

E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148

5 0
3 years ago
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