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blagie [28]
3 years ago
5

Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. a) What

is the probability that we obtain a sum of 3 before we obtain a sum of 7? Hint: Let A denote the event of a sum of 3. Let B denote the event of no sum of 3 or 7. Now, begin to list the out the sample space in terms of A and B - it should be recognizable in the lecture notes. Hint: A useful infinite series result when |z| < 1: 1 + z + z 2 + z 3 + z 4 + . . . = 1 1 − z
Mathematics
1 answer:
Marrrta [24]3 years ago
5 0
رتلمفخيحاحتحثحقنغنفهصنفوامكغمبخصخا
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7×300=7× blank hundreds
VladimirAG [237]
7x300=7x 3 hundres because in 300 hundreds there is 3 hundreds.I know that didnt make sense.so basically just look at the front number number its 3 so 3 hundreds but if your working with thousands and the front number is 5 its 5 thousands.I hope this was helpful.
6 0
4 years ago
Which is equivalent to 80 1/4x
Illusion [34]

Answer:

20x

Step-by-step explanation:

You would multiply 1/4 times 8. Or instead, you could divide 80 by 4. You would get 20 then you would multiply that by x and get 20x. Hope this helps.

6 0
3 years ago
Solve the initial-value problem<br><br> y' = x^4 - \frac{1}{x}y, y(1) = 1.
natta225 [31]

The ODE is linear:

y'=x^4-\dfrac yx

y'+\dfrac yx=x^4

Multiplying both sides by x gives

xy'+y=x^5

Notice that the left side can be condensed as the derivative of a product:

(xy)'=x^5

Integrating both sides with respect to x yields

xy=\dfrac{x^6}6+C

\implies y(x)=\dfrac{x^5}6+\dfrac Cx

Since y(1)=1,

1=\dfrac16+C\implies C=\dfrac56

so that

\boxed{y(x)=\dfrac{x^5}6+\dfrac5{6x}}

4 0
3 years ago
Consider the sequence: -7, -2, 3, 8…. What is f(7)?
wlad13 [49]
You are adding by 5 in the sequence.
If f(1) is -7
f(2) = -2
f(3)= 3
f(4)=8
f(5)=13
f(6)=18
Therefore, f(7)=23
5 0
3 years ago
How many tons are in 16000
Svetach [21]

Answer: 8 tons

Step-by-step explanation:

7 0
3 years ago
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