(2/1 )(3/2)(4/3)...(n-1/n-2)(n/n-1)(n+1/n) = n + 1
n + 1 = n + 1
1 = 1 ( Fact )
Thus it is not importent what the value of n is at all because the equation is always have the equal number ( which is 1 ) in both sides .
Answer:
a) 0
b) 4.5
c) 3
d) 6
Step-by-step explanation:
Formula: 3x + 4y = 12
The questions contains the values of x. So here's an example of:
d) x = -4, y = ?
Using the formula: 3(-4) + 4y = 12
-12 + 4y = 12
4y = 24
y = 6
Do the same for the others!
Let's begin by assigning a letter to represent our unknown: x = # of toppings to be added to each pizza Now let's come up with expressions for the cost of each pizza with its toppings: Palanxio's Pizza: 6.80 + 0.90x Guido's Pizza: 7.30 + 0.65x Now let's equate the costs of both pizzas and solve for our unknown (x): 6.80 + 0.90x = 7.30 + 0.65x 0.90x - 0.65x = 7.30 - 6.80 0.25x = 0.50 x = 0.50/0.25 x = 2 toppings to be added to each pizza Finally, we can verify our answer by plugging it back into our equation and see if both costs are the same: 6.80 + 0.90x = 7.30 + 0.65x [equation] 6.80 + 0.90(2) = 7.30 + 0.65(2) [answer plug-in] 6.80 + 1.80 = 7.30 + 1.30 $8.60 = $8.60 [both costs are the same] Since both costs are the same, we are confident that our answer is correct.
9514 1404 393
Answer:
Step-by-step explanation:
The applicable rules of exponents are ...
(a^b)(a^c) = a^(b+c)
a^0 = 1 . . . . for a ≠ 0
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And, it is convenient to know the cubes of small integers:
1³ = 1; 2³ = 8; 3³ = 27; 4³ = 64; 5³ = 125
6³ = 216; 7³ = 343; 8³ = 512; 9³ = 729; 10³ = 1000
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1) p^3 × p^5 = p^-12 × p^y
Equating exponents:
3 + 5 = -12 + y
20 = y . . . . . . . . . . . add 12
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2) 64 × 4^5 = 4^3 × 4^5 = 4^(3+5) = 4^8
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3) 10^x = 1 = 10^0
x = 0
Answer:
And we can solve for the value of x and we got:
![22.1 -[1*0.1 +10*0.1 +15*0.1 + 20*0.1 +25*0.1 ] = 0.5 x](https://tex.z-dn.net/?f=%2022.1%20-%5B1%2A0.1%20%2B10%2A0.1%20%2B15%2A0.1%20%2B%2020%2A0.1%20%2B25%2A0.1%20%5D%20%3D%200.5%20x)
Step-by-step explanation:
For this case we have the following distribution:
X 1 10 15 20 25 x
P(X) 0.1 0.1 0.1 0.1 0.1 0.5
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
The expected value is given by:
And replacing we have this:
And we can solve for the value of x and we got:
