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Ymorist [56]
3 years ago
6

Find m please help me i’ve been stuck on this problem for 30 minutes!!!

Mathematics
1 answer:
victus00 [196]3 years ago
8 0

6×+8=14 5×+2=7 14×7=98 ∆m=4.66

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36 is 0.09% of what number
alexgriva [62]
Х-100%
36-0,06%
x=(36*100)\0,06=60000
Answer:60000
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What is 1% of $26 I gotta get this in before Thursday
Aliun [14]

Answer:

.26

Step-by-step explanation:

(.01)(26)=.26

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3 years ago
The image of the point
Stella [2.4K]

Answer:

(2, 0)

Step-by-step explanation:

First, find how the other point was transformed:

The x coordinate increased by 1, and the y coordinate increased by 4.

Now, do these same transformations to the other point, (1, -4)

x coordinate: 1 + 1 = 2

y coordinate: -4 + 4 = 0

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8 0
4 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
What are the ratios?​
Ilya [14]

Answer:

3/5

Step-by-step explanation:

5 0
3 years ago
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