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Andrews [41]
3 years ago
6

How do you solve 100^(5x+2)=1000^(4x-1)?

Mathematics
1 answer:
katrin [286]3 years ago
3 0
This whole equation is also equal to 10^10(5x+2)=10^100(4x-1). From there you can divide both sides by 10 and distribute on both sides. From there, it’s just basic algebra
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A factory ships widgets in crates. There are 12 boxes in each crate. Each box holds 275 widgets. How many widgets are in one cra
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There are 3300 widgets in each crate.
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Hope this helps,
~ Harley Quinn ~
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Which is true of a geometric straight line?
bazaltina [42]

Answer:

D) no dimensions, only position.

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4 0
4 years ago
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Determine whether the function f(x) -1 / x^2 + x^4 is even odd or neither
kirill [66]

If f(-x) = f(x) then the function is even.

If f(-x) = -f(x) then the function is odd.

We have: f(x)=\dfrac{-1}{x^2+x^4}

calculate f(-x):

f(-x)=\dfrac{-1}{(-x)^2+(-x)^4}=\dfrac{-1}{(-1x)^2+(-1x)^4}\\\\=\dfrac{-1}{(-1)^2x^2+(-1)^4x^4}=\dfrac{-1}{1x^2+1x^4}=\dfrac{-1}{x^2+x^4}=f(x)

f(-x) = f(x), therefore your answer: The function f(x) is even.

If we have: f(x)=\dfrac{-1}{x^2}+x^4

f(-x)=\dfrac{-1}{(-x)^2}+(-x)^4=\dfrac{-1}{x^2}+x^4

f(-x) = f(x)

f(x) is even

5 0
3 years ago
Let X equal the number of hours that a randomly selected person from City A reads the news online each day. Suppose that X has a
Nina [5.8K]

Answer:

P(Y>X) = \frac{17}{32}

Step-by-step explanation:

Recall that since X is uniformly distributed over the set [1,4] we have that the pdf of X is given by f(x) = \frac{1}{3} if 1\leq x \leq 4 and 0 otherwise. In the same manner, the pdf of Y is given by g(y) = \frac{1}{4} if 1\leq y\leq 5 and 0 otherwise.

Note that if Y is in the interval (4,5] then Y>X by default. So, in this case we have that P(Y>X| y in (4,5]) = 1. We want to calculate the probability of having Y in that interval . That is

P(Y>4) = \int_{4}^{5}\frac{1}{4}dy = \frac{1}{4}. Thus, P(Y\leq 4 ) = 1 - P(Y>4)= \frac{3}{4}.

We want to proceed as follows. Using the total probability theorem, given two events A, B we have that

P(A) = P(A|B)\cdot P(B) + P(A|B^c) \cdot P(B^c) In this case, A is the event that Y>X and B is the event that Y is in the interval (4,5].

If we assume that X and Y are independent, then we have that the joint pdf of X,Y is given by h(x,y) = f(x)g(y) = \frac{1}{12} when 1\leq x \leq 4, 1\leq y \leq 4. We can draw the region were Y>X and the function h(x,y) is different from 0. (The drawing is attached). This region is described as follows: 1\leq x \leq 4 and x\leq y \leq 4, then (the specifics of the calculations of the integrals are ommitted)

P(Y>X| y \in [1,4)) = \int_{1}^{4}\int_{x}^{4}\frac{1}{12}dy dx = \frac{1}{12}\int_{1}^{4} (4-x) dx = \frac{1}{12}\left.(4x-\frac{x}{2})\right|_{1}^{4}= \frac{1}{12}(4\cdot 4 - \frac{4^2}{2}-(4-\frac{1}{2}) = \frac{9}{2\cdot 12} = \frac{3}{8}

Thus,

P(Y>X) = 1\cdot \frac{1}{4} + \frac{3}{8}\cdot \frac{3}{4} = \frac{17}{32}

5 0
4 years ago
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