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Zina [86]
3 years ago
8

Simply the expression below . 8+ ( 15 - 3 ) + 4 help please .

Mathematics
2 answers:
Alona [7]3 years ago
7 0
8 + (15-3) + 4
8 + 12 + 4
20 + 4
24
your answer is 24
maxonik [38]3 years ago
7 0

Answer:

24

Step-by-step explanation:

8+ ( 15 - 3 ) + 4

8+12+4

20+4

24

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An automobile with 0.279 m radius tires travels 65000 km before wearing them out. How many revolutions do the tires make, neglec
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Revolutions=37079108.61 rev

Step-by-step explanation:

First if all we have to make the units same to calculate the revolutions

Since distance gis gien in km and radius is given in m we will covert distance to m first.

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3 years ago
Samantha answered 90% of the questions on her math test and 80 % of the questions on her history test correctly. If each test ha
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Read 2 more answers
(1 point) An insurance company offers its policyholders a number of different premium payment options. For a randomly selected p
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Answer:

a)

X    |  1        3      5       7

f(X) | 0.4   0.2   0.2    0.2

b) P(4

Step-by-step explanation:

For this case we have defined the cumulative distribution function like this:

F(X) = 0, x

F(X) = 0.4, 1 \leq x

F(X) = 0.6, 3 \leq x

F(X) = 0.8, 5 \leq x

F(X) = 1, x \geq 7

And we know that the general definition for the distribution function is given by:

F(x) = P(X \leq x) = \sum_{i\leq k} f(i)

Where f represent the density function.

Part a

For this case we need to find the density function, so we can find the values for the density for each value of X = 1,2,3,4,5,6,7,... since X is a discrete random variable.

f(1) = P(X=1) = P(X \leq 1) - P(X=0) = F(1) -F(0) = 0.4-0=0.4

f(2) = P(X=2) = P(X \leq 2) - P(X=0)- P(X=1) = F(2) -F(1) = 0.4-0.4=0

f(3) = P(X=3) = P(X \leq 3) - P(X=0)- P(X=1) -P(X=2) = F(3) -F(2) = 0.6-0.4=0.2

f(4) = P(X=4) = P(X \leq 4) - P(X=0)- P(X=1) -P(X=2)-P(X=3) = F(4) -F(3) = 0.6-0.6=0

f(5) = P(X=5) = P(X \leq 5) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4) = F(5) -F(4) = 0.8-0.6=0.2

f(6) = P(X=6) = P(X \leq 6) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4)-P(X=5) = F(6) -F(5) = 0.8-0.8=0

f(7) = P(X=7) = P(X \leq 7) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4)-P(X=5)-P(X=6) = F(7) -F(6) = 1-0.8=0.2

And for any value higher than 7 we have that:

x_i \in [8,9,10,...]

f(x_i) = F(X_i) -F(X_i -1) = 1-1=0

So then we have our density function defined like this:

X    |  1        3      5       7

f(X) | 0.4   0.2   0.2    0.2

Part b

For this case we want to find this probability P(4

And since the random variable is discrete we can write this like that:

P(4

5 0
3 years ago
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