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Nat2105 [25]
2 years ago
5

Please help me understand how to solve this

Mathematics
1 answer:
ss7ja [257]2 years ago
6 0
Ok, Soooo.... It is all about probability. First take down all your info. 

A die has 6 different sides. 2 die = 12 sides. 2 different ages. The chances of one of their ages coming up on the first throw is very unlikely, because of all the different combinations you could come up with. Take that into consideration. 
Now, you would add up all your information. My guess is that they want to see how you work out the question to see how soon their age would come. Note down that there are many different options. 

Hope this helps, 
Please Rate as Best Answer
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WORTH 50! WILL AWARD BRAINLIEST! PLEASE SHOW STEPS! THANK YOU!
yanalaym [24]

Answer:

x = 45

Step-by-step explanation:

Re-order terms so constants are on the left
15 * x/9 - 42 =28

Combine multiplied terms into a single fraction
15x/9 - 42 = 28

Cancel terms that are in both the numerator and denominator
5x/3 - 42 = 28


4 0
2 years ago
The ratio of the heights of two similar cylinders is 1:3. If the volume of the smaller cylinder is 67cm cubed, find the volume o
irina1246 [14]

Answer:

volume = 1809 cm³

Step-by-step explanation:

The ratio of the height of two similar cylinder is 1:3 . The volume of the smaller cylinder is 67 cm³. The volume of the larger cylinder can be computed below:

If 2 object are similar and the ratio of the corresponding sides is x ,then the ratio of their volume is x³.

ratio of height = 1/3

ratio of volume = (1/3)³ = 1/27

volume of smaller cylinder/volume of larger cylinder = 1/27 = 67/x

where

x = volume of the larger cylinder

x = 27 × 67

x = 1809  cm³

volume = 1809 cm³

5 0
3 years ago
An item is regularly priced at $90. Is is on Sale for 20% off the regular price. How much (in dollars) is discounted from the re
fomenos

Answer:

18

Step-by-step explanation:

because 20% of 90 is 18

7 0
2 years ago
Read 2 more answers
Branliest offered<br> solve the proportion<br> 12/x=6/7<br> 6x/4=8/12<br> 7/x+13=4/12<br> y+5/y=10/8
Tju [1.3M]
12/x=6/7 answer: x=14
6x/4=8/12 answer x = 0.4444444444
7/x+13=4/12 answer x=8
y+5/y=10/8 answer y=20
6 0
3 years ago
9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operat
Marysya12 [62]

Answer:

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

n = 2000, p = 1-0.995 = 0.005

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

P(X < 5) + P(X \geq 5) = 1

We want P(X \geq 5)

So

P(X \geq 5) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044

P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445

P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235

P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480

P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765

P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

4 0
3 years ago
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