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AveGali [126]
3 years ago
8

If any one wana talk to me ill give you mu tomy meting

Mathematics
2 answers:
AURORKA [14]3 years ago
8 0

Answer:

iteee

Step-by-step explanation:

bulgar [2K]3 years ago
7 0

Answer:

Um, i think that's against the rules of brainly but I don't even know for sure.

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What are the factors of 36?
denis23 [38]

Answer:

1, 2, 3, 4, 6, 9, 12, 18, 36

Step-by-step explanation:

3 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
What’s the missing factor?
Vesna [10]

The missing factor is 1 5/8

<u>Step-by-step explanation:</u>

Let us consider the missing factor has 'x'

The given expression is 4 1/2 - x = 2 7/8

<u>To find x :</u>

The first step is to bring all the constant terms on on side and x on the other side,

⇒ 4 1/2 - 2 7/8 = x

Now, the numbers are in improper fraction (mixed fraction). So you have to convert them into proper fraction.

To convert the improper fraction 4 1/2 into proper fraction,

Multiply the denominator 2 with 4 and the result 8 is added with the numerator 1.

⇒ 4 1/2 = 9/2.

To convert the improper fraction 2 7/8 into proper fraction,

Multiply the denominator 8 with 2 and the result 16 is added with the numerator 7.

⇒ 2 7/8 = 23/8.

Now replace them in the given expression. The expression becomes as :

⇒ (9/2) - (23/8) = x

Cross multiply 2 and 8,

⇒ (72 - 46) / 16 = x

⇒ 26 / 16 = x

⇒ 13/8 = x

The x value is 13/8. You have to convert it into mixed fraction (since the question is given in the form of mixed fraction).

<u>To convert into mixed fraction,</u>

Divide 13 by 8 and write them in the form of Quotient (Reminder / Divisor).

⇒ 13 ÷ 8

⇒ 1 (quotient)

⇒ 5 (remainder)

The divisor is 8.

Therefore, x = 13/8 can be written as 1 5/8.

The missing factor is 1 5/8.

8 0
4 years ago
I need help with my homework this is hard pls help thx!<br> z=m-x solve for x
lisov135 [29]

Answer:

x = m - z

Step-by-step explanation:

z = m - x       add x to both sides

z + x = m - x + x

z + x = m            subtract z from both sides

z - z + x = m - z

x = m - z

6 0
3 years ago
Read 2 more answers
I am offering another 100 points
Ivahew [28]

Answer:

\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.

Step-by-step explanation:

<u>Perfect squares</u>: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ...

To find \sf \sqrt{75} , identify the perfect squares immediately <u>before</u> and <u>after</u> 75:

  • 64 and 81

\begin{aligned}\sf As\;\; 64 < 75 < 81\; & \implies \sf \sqrt{64} < \sqrt{75} < \sqrt{81}\\&\implies \sf \;\;\;\;\;8 < \sqrt{75} < 9 \end{aligned}

\sf Since \;\sqrt{\boxed{64}}=\boxed{8}\;and\;\sqrt{\boxed{81}}=\boxed{9}\; \textsf{it is known that $\sqrt{75}$ is between}\\\\\sf \boxed{8}\;and\;\boxed{9}\;.

See the attachment for the correct placement of \sf \sqrt{75} on the number line.

6 0
1 year ago
Read 2 more answers
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