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3 years ago

Use the Fundamental Counting Principle to find the total number of outcomes in each situation.

Mathematics

2 answers:

Dmitrij [34]3 years ago

6 0

I think it’s B or it’s either C

bija089 [108]3 years ago

3 0

When in doubt pick c lol

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(please help.) (I screenshot the question and choices )

padilas [110]

The second answer on the list is the correct one.

8 0

3 years ago

ILL MARK BRAINIEST IF YOU DO THIS RIGHT!!! DO 9 AND 10!!!

yuradex [85]

Number 9 is L is greater than or equal to 110

4 0

3 years ago

Simplify a•a²•a³ of the following

professor190 [17]

Answer:

a^6

Step-by-step explanation:

when multiplying exponents, add the exponents together. you can distribute to prove it, a^2 is a•a, a^3 is a•a•a so by bringing it together you get a•(a•a)•(a•a•a) which gets a 6 times or a^6

5 0

3 years ago

A system has two failure modes. One failure mode, due to external conditions, has a constant failure rate of 0.07 failures per y

nadya68 [22]

Answer:

0.9177

Step-by-step explanation:

let us first represent the two failure modes with respect to time as follows

R₁(t) for external conditions

R₂(t) for wear out condition ( Wiebull )

Now,

$R1(t) = e^{-nt} .....1$

where t = time in years = 1,

n = failure rate constant = 0.07

Also,

$R2(t)=e^{-(\frac{t}{Q} )^{B} }......2$

where t = time in years = 1

where Q = characteristic life in years = 10

and B = the shape parameter = 1.8

Substituting values into equation 1

$R1(t) = e^{-(0.07)(1)} \\\\R1(t) = e^{-0.07}$

Substituting values into equation 2

$R2(t)=e^{-(\frac{1}{10} )^{1.8} }\\\\R2(t)=e^{-(0.1)}^{1.8} }\\\\R2(t)=e^{-0.0158}$

let the <em>system reliability </em>for a design life of one year be Rs(t)

hence,

Rs(t) = R1(t) * R2(t)

t = 1

$Rs(1) = [e^{-0.07} ] * [e^{-0.0158} ] = 0.917713$

Rs(1) = 0.9177 (approx to four decimal places)

5 0

3 years ago

What is the Divisor for equivalent fraction ?<br> 4/12=1/3

Lorico [155]

Do do this you divide the largest denominator by the smallest;

12/3 = 4

The answer is 4

Hope this helps! :)

5 0

3 years ago