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mamaluj [8]
4 years ago
10

I’m so bad at these problems so please help ): would it also be mm or square mm

Mathematics
2 answers:
LekaFEV [45]4 years ago
6 0
My answer would be square mm
Fittoniya [83]4 years ago
6 0
It would be square mm
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Diane Van Os decided to buy a new car since her credit union was offering such low interest rates. She borrowed 30,000 at 4.75%
mixer [17]

Answer:

Total Interest Diane has to pay is $1,745

Step-by-step explanation:

<u>Behind the Excel sheet</u>

Total Number of Days:

It was the sum 2016 + 2017 + 3 months of 2018

                             19 + 365   + 31 + 28 + 4

<u>Exact Intrest on Loan</u>

the Borrowed Amount is multiplied by Exact Intrest Rate multiplied by Total Number of Days divided by Number of Days in a Year

$30,000 x 0.0475 x (447/365)

4 0
2 years ago
What’s the answer to this?v
koban [17]
I think it’s A because it’s the only one that makes sense
3 0
3 years ago
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What is the start time?<br><br> End time:10:08<br> elapsed time :30 minutes
Hoochie [10]

Answer:

9:38

Step-by-step explanation:

BRAINLY PATROL DONT DELETE THIS!

10:08-30=9:38

5 0
2 years ago
Jacksons kitty weighed 2 pounds and 3 ounces a month later the kitty weighed 56 ounces how much weight did the kitty gain in tha
8_murik_8 [283]

Answer:

21 ounces

Step-by-step explanation:

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3 years ago
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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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