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USPshnik [31]
3 years ago
10

Round this “0.7660444431”into 4 decimals places!!!!!

Mathematics
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

.7660

Step-by-step explanation:

when you get to 0 at 7660 you round with the number next to it if its 5 or bigger you add a one to the number since 4 isn't greater than or equal to 5 than your numbers stay the same

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Suppose that a diamond weighs 0.0182 carat.What is that number in expanded form
kvv77 [185]
0.0182=(0 x 1)+ (0/10) + (1/100) + (8/1000) + (2/10000).
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3 years ago
The matrix 1 2 4(0 1 6/7 is being used to solve a system of equations. The solution to the system is
loris [4]
| 1 2 ||x| = |4| - | 3 -1 ||y| |6| <span>-3 | 1 2||x| = |4| </span>
<span>| 3 -1||y| |6| </span>

<span>|-3 -6||x| = |-12| </span>
<span>| 3 -1||y| |6 | </span>


<span>| 0 -7 ||x| = | -6 | </span>
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5 0
3 years ago
6. Lily and Sara each had an equal amount of money at first. After Lily
kiruha [24]
X-18 = 2 (X-25)
x-18 = 2x - 50
-18 = x - 50
32 = x

check
32-18 = 14 = 2(32-25) = 2*7
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8 0
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Least two digit number exactly divisible by 6<br>​
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Step-by-step explanation:

8 0
2 years ago
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Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab
Sphinxa [80]

Answer:  a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Step-by-step explanation:

Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:

Since we have given that

\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5

Using the poisson process, we get that

(a) less than or equal to 12

probability=  P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980

(b) equal to 13

probability= P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013

(c) greater than 12

probability= P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020

(d) equal to 20

probability= P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026

(e) between 10 and 15, inclusively

probability=P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318

Hence, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

6 0
3 years ago
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