Which values from this set {−30, −20, −10, 0, 10, 20} satisfy this inequality?
1 answer:
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The answer is A: x=1/2 , y= 1/3
Solve the following system:
{2 x + 3 y = 2 | (equation 1)
{6 x + 18 y = 9 | (equation 2)
Swap equation 1 with equation 2:
{6 x + 18 y = 9 | (equation 1)
{2 x + 3 y = 2 | (equation 2)
Subtract 1/3 × (equation 1) from equation 2:
{6 x + 18 y = 9 | (equation 1)
{0 x - 3 y = -1 | (equation 2)
Divide equation 1 by 3:
{2 x + 6 y = 3 | (equation 1)
{0 x - 3 y = -1 | (equation 2)
Multiply equation 2 by -1:
{2 x + 6 y = 3 | (equation 1)
{0 x+3 y = 1 | (equation 2)
Divide equation 2 by 3:
{2 x + 6 y = 3 | (equation 1)
{0 x+y = 1/3 | (equation 2)
Subtract 6 × (equation 2) from equation 1:
{2 x+0 y = 1 | (equation 1)
v0 x+y = 1/3 | (equation 2)
Divide equation 1 by 2:
{x+0 y = 1/2 | (equation 1)
v0 x+y = 1/3 | (equation 2)
Collect results:
Answer: {x = 1/2, y = 1/3
Solve the following system:
{2 x + 3 y = 2 | (equation 1)
{6 x + 18 y = 9 | (equation 2)
Swap equation 1 with equation 2:
{6 x + 18 y = 9 | (equation 1)
{2 x + 3 y = 2 | (equation 2)
Subtract 1/3 × (equation 1) from equation 2:
{6 x + 18 y = 9 | (equation 1)
{0 x - 3 y = -1 | (equation 2)
Divide equation 1 by 3:
{2 x + 6 y = 3 | (equation 1)
{0 x - 3 y = -1 | (equation 2)
Multiply equation 2 by -1:
{2 x + 6 y = 3 | (equation 1)
{0 x+3 y = 1 | (equation 2)
Divide equation 2 by 3:
{2 x + 6 y = 3 | (equation 1)
{0 x+y = 1/3 | (equation 2)
Subtract 6 × (equation 2) from equation 1:
{2 x+0 y = 1 | (equation 1)
v0 x+y = 1/3 | (equation 2)
Divide equation 1 by 2:
{x+0 y = 1/2 | (equation 1)
v0 x+y = 1/3 | (equation 2)
Collect results:
Answer: {x = 1/2, y = 1/3
Answer:
The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.
Step-by-step explanation:
Let <em>X</em> = number of students who have laptops.
The probability of a student having a laptop is, P (X) = <em>p</em> = 0.37.
A random sample of <em>n</em> = 30 students is selected.
The event of a student having a laptop is independent of the other students.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The mean and standard deviation of a binomial random variable <em>X</em> are:
Compute the mean of the random variable <em>X</em> as follows:
The mean of the random variable <em>X</em> is 1.11.
Compute the standard deviation of the random variable <em>X</em> as follows:
The standard deviation of the random variable <em>X</em> is 0.836.