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DedPeter [7]
3 years ago
14

WILL MARK BRAINLIEST IF CORRECT! :)

Mathematics
1 answer:
Oduvanchick [21]3 years ago
8 0

Answer:

probability is 1/6

Step-by-step explanation:

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Find the volume of a cylinder that has a diameter of 12 in. and a height of 15 in
daser333 [38]

Answer:

<h3>\sf{ \boxed{ \bold{1697.14}}}</h3>

Step-by-step explanation:

Given,

diameter ( d ) = 12 in

height ( h ) = 15 in

<u>finding </u><u>the </u><u>radius </u><u>of </u><u>a </u><u>cylinder</u>

Radius is just half of diameter.

Radius ( r ) = 12 / 2 = 6 in

<u>finding </u><u>the </u><u>volume </u><u>of </u><u>a </u><u>cylinder </u><u>having </u><u>radius </u><u>of </u><u>6</u><u> </u><u>in </u><u>and </u><u>height </u><u>of </u><u>1</u><u>5</u><u> </u><u>in</u>

Volume of a cylinder = <u>\sf{\pi  \:  {r}^{2} h}</u>

⇒\sf{ \frac{22}{7}  \times  {6}^{2}  \times 15}

⇒\sf{ \frac{22}{7}  \times  36 \times 15}

⇒<u>\sf{1697.14} \: in</u>

Hope I helped!

Best regards!!

6 0
4 years ago
Read 2 more answers
Figure out this number pattern:<br>1,5,14,30,55
Amanda [17]

5 - 1 = 4 = 2^2

14 - 5 = 9 = 3^2

30 - 14 = 16 = 4^2

55 - 30 = 25 = 5^2

Start with 1, then add 2^2, 3^3, 4^2, 5^2, 6^2, etc.

5 0
4 years ago
Bit transmission errors between computers sometimes occur, where one computer sends a 0 but the other computer receives a 1 (or
lana66690 [7]

Answer:

(a) Probability that a triplet is decoded incorrectly by the receiving computer. = 0.010

(b)

(1 – p) = 0.010

(c)

E(x) = 25000 x 0.010

     = 259.2

Explanation has given below.

Step-by-step explanation:

Solution:

(a) Probability that a triplet is decoded.

2 out of three

P = 0.94, n = 3

m= no of correct bits

m   bit (3, 0.94)

At p(m≤1) = B (1; 3, 0.94)

 = 0.010

(b) Using your answer to part (a),

(1 – p) = 0.010

Error for 1 bit transmission error.

(c)  How does your answer to part (a) change if each bit is repeated five times (instead of three?

P( m ≤ 2 )

L = Bit (5, 0.94)

   = B (2; 5, 0.94)

   = 0.002

(d)  Imagine a 25 kilobit message (i.e., one requiring 25,000 bits to send). What is the expected number of errors if there is no bit repetition implemented? If each bit is repeated three times?

Given:

h = 25000

Bits are switched during transmission between two computers = 6% = 0.06

m = Bit (25000, 0.06)

E(m) = np

        = 25000 x 0.06

         = 1500

m = Bit (25000, 0.01)

E(m) = 25000 x 0.010

     = 259.2

3 0
3 years ago
Cos (90 - x) = -0.2. What is sin(x)?
Aloiza [94]

Answer:

-0.2

Step-by-step explanation:

Sine and cosine are co-functions.

That means, cos(90-x)=sin(x) or sin(90-x)=cos(x).

So here since cos(90-x)=-0.2, then sin(x)=-0.2.

4 0
3 years ago
Read 2 more answers
Suppose the lifetime, in years, of a motherboard is modeled by a Gamma distribution with parameters α=80 and λ=4. Use the Centra
timurjin [86]

Answer:

Therefore there's a 99.99% probability the motherboard of your new computer will last for at least 15 years.

Step-by-step explanation:

This is the general idea to solve the problem.

Suppose that the mean and variance of the your distribution are .\mu , \sigma respectively. Then, according to the problem you are looking for the probability.

P(X \geq 15) = 1 - P(X

Consider then the following random variable.

T  =   X_1 + X_2 + .... + X_{14}

Using the central limit theorem  T distribution will be close to normal, and its mean and variance will be  14\mu , 14\sigma  , respectively. Therefore you just have to find the probability that a normally distributed  random variable with that mean and that variance which I just mentioned is less than 14.

For this case we have that

\mu = \alpha / \gamma  = 20 \\\sigma = \alpha(\alpha+1) / \gamma^2  -  (\alpha/ \gamma)^2  = 5

Then you have that

14\mu = 280\\14\sigma = 70\\

and we have that if  N is a normally distributed random variables with mean 280 and variance 70 we have that

P(N \leq 14 ) = 0.0001

the actual probability we are looking for is

1-P(N\leq14) =  1-(0.0001) = 0.9999

Therefore there's a 99.99% probability the motherboard of your new computer will last for at least 15 years.

5 0
3 years ago
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