Answer: 90% confidence interval is; ( - 0.0516, 0.3752 )
Step-by-step explanation:
Given the data in the question;
n1 = 72, n2 = 17
P1 = 54 / 72 = 0.75
P2 = 10 / 17 = 0.5882
so
P_good = 0.75
P_bad = 0.5882
standard ERRROR will be;
SE = √[(0.75×(1-0.75)/72) + (0.5882×(1-0.5882)/17)]
SE = √( 0.002604 + 0.01424)
SE = 0.12978
given confidence interval = 90%
significance level a = (1 - 90/100) = 0.1, |Z( 0.1/2=0.05)| = 1.645 { from standard normal table}
so
93% CI is;
(0.75 - 0.5882) - 1.645×0.12978 <P_good - P_bad< (0.75 - 0.5882) + 1.645×0.12978
⇒0.1618 - 0.2134 <P_good - P_bad< 0.1618 + 0.2134
⇒ - 0.0516 <P_good - P_bad< 0.3752
Therefore 90% confidence interval is; ( - 0.0516, 0.3752 )
Answer:
A) x=39
Step-by-step explanation:
117 = 3x
117 divided by 3 = 39
3x = 117
3(39) = 117
117 = 117
A) x = 39
The factors of polynomial are (x-1) and (x+3)
Step-by-step explanation:
If (x+9) is a factor of P(x)= x³+11x²+15x-27 then the P(x) must be completely divisible by (x+9)
After finding the polynomial, we need to factorise the polynomial by utilising the quadratic equation.
All the process has been described in image attached-
S₁, S₂, S₃ represents various steps involved in solving the problem.
It is found that (x+9) is a factor of x³+11x²+15x-27
The polynomial found after dividing x³+11x²+15x-27 by (x+9) is x²+2x-3
Which can be further factorised by quadratic rule into the factors as (x-1) and (x+3)
Hence the factors are (x-1) and (x+3)
For this case we have the following system of equations:
5x + 3y = 17
-8x - 3y = 9
The solution to this system can be written in matrix form as:
Ax = b
Where, A is a matrix that is given by:
A = [5 3; -8 -3]
The determinant of this matrix 2 * 2 is given by:
lAl = (5) * (- 3) - (3) * (- 8)
lAl = -15 - (-24)
lAl = -15 + 24
lAl = 9
Answer:
The determinant for solving this linear system is:
lAl = 9
Answer: Option A is correct.
Explanation: Both fractions equal 6/5 when fully simplified. Questions like these are easy to check. All you have to do is divide 18 by 15 and divide 24 by 20. Both will equal 1.2.
