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Olin [163]
3 years ago
8

4 1/3 · 3/5 express your answer in simplest form. A. 2 3/5 B. 2 C. 4 1/5 D. 1 1/15

Mathematics
2 answers:
a_sh-v [17]3 years ago
7 0

Oh the answer is a I looked it up this time

Vesna [10]3 years ago
3 0

Find common denominators:

4 1/3 = 4 5/15 = 65/15

3/5 = 9/15

65/15 x 9/15 = 2.6 = 2 6/10 = 2 3/5

Final Answer:

A. 2 3/5

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Help photo question. Mathematics. :D
Aleks04 [339]
M = 4
n = 1

If you substitute in your values, you can see that 4 + 1 = 5
And 4 - 1 = 3
Hope this helped :)
5 0
3 years ago
Let X denote the data transfer time (ms) in a grid computing system (the time required for data transfer) between a "worker" com
My name is Ann [436]

Answer:

a. The value of alpha is 3.014 and the value of beta is 12.442

b. The probability that data transfer time exceeds 50ms is 0.238

c. The probability that data transfer time is between 50 and 75 ms is 0.176

Step-by-step explanation:

a. According to the given data we have that the mean and standard deviation of the random variable X are 37.5 ms and 21.6.

Therefore, E(X)=37.5 and V(X)=(21.6)∧2  

To calculate alpha we would have to use the following formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To calculate beta we would have to use the following formula:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. E(X)=37.5 and V(X)=(21.6)∧2  

Therefore, P(X>50)=1−P(X≤50)

Hence, To calculate the probability that data transfer time exceeds 50ms we use the following formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The probability that data transfer time exceeds 50ms is 0.238

c. E(X)=37.5 and V(X)=(21.6)∧2  

​Therefore, P(50<X<75)=P(X<75)−P(X<50)  

Hence, To calculate the probability that data transfer time is between 50 and 75 ms we use the following formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time is between 50 and 75 ms is 0.176

6 0
3 years ago
Manny is covering a book. The front of the book is 11.2 inches high and 7.3 inches wide. What is the area of the front of the bo
irina [24]

Answer:

81.76in^2

Step-by-step explanation:

Given data

Lenght of book=  11.2 inches

WIdth of book= 7.3 inches

We know that Area= Length* Width

Area= 11.2*7.3

Area= 81.76 in^2

Hence the area of the front of the book is is 81.76in^2

7 0
3 years ago
During a solar eclipse, what is causing the earth’s view of the sun to be darkened?
Serggg [28]

Answer:

the moon blocks light from the sun by coming between the earth and the sun

7 0
2 years ago
Read 2 more answers
5c squared- d squared+3/2c - 4d...c= -1. d= -4​
riadik2000 [5.3K]

For this case we have the following expression:

5c ^ 2-d ^ 2 + \frac {3} {2} c-4d

We must find the value of the expression when:

c = -1\\d = -4

Substituting we have:

5 (-1) ^ 2 - (- 4) ^ 2 + \frac {3} {2} (- 1) -4 (-4) =\\5 (1) -16- \frac {3} {2} + 16 =\\5-16- \frac {3} {2} + 16 =\\-11- \frac {3} {2} + 16 =\\- \frac {25} {2} + 16 =\\\frac {-25 + 32} {2} =\\\frac {7} {2}

Finally, the value of the expression is:

\frac {7} {2}

ANswer:

\frac {7} {2}

5 0
3 years ago
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