Answer:19
Step-by-step explanation:
7x3=21
21-2=19
Answer:
- 4x² - 13x + 8 = 0
- 4x² - 11x + 5 = 0
- 16x² - 41x + 1 = 0
- x² + 5x + 4 = 0
- x² - 66x + 64 = 0
Step-by-step explanation:
<u>Given</u>
- α and β are roots of 4x²-5x-1=0
<u>Then the sum and product of the roots are:</u>
- α+b = -(-5)/4 = 5/4
- αβ = -1/4
(i) <u>Roots are α + 1 and β + 1, then we have:</u>
- (x - (α + 1))(x - (β + 1)) = 0
- (x - α - 1)(x - β - 1) = 0
- x² - (α+β+2)x + α+β+ αβ + 1 = 0
- x² - (5/4+2)x +5/4 - 1/4 + 1 = 0
- x² - 13/4x + 2= 0
- 4x² - 13x + 8 = 0
(ii) <u>Roots are 2 - α and 2 - β, then we have:</u>
- (x + α - 2)(x + β - 2) = 0
- x² + (a + β - 4)x - 2(α + β) + αβ + 4 = 0
- x² + (5/4 - 4)x - 2(5/4) - 1/4 + 4 = 0
- x² - 11/4x - 10/4 - 1/4 + 16/4 = 0
- x² - 11/4x + 5/4x = 0
- 4x² - 11x + 5 = 0
(iii) <u>Roots are α² and β², then:</u>
- (x - α²)(x-β²) = 0
- x² -(α²+β²)x + (αβ)² = 0
- x² - ((α+β)² - 2αβ)x + (-1/4)² = 0
- x² - ((5/4)² -2(-1/4))x + 1/16 = 0
- x² - ( 25/16 + 1/2)x + 1/16 = 0
- x² - 33/16x + 1/16 = 0
- 16x² - 33x + 1 = 0
(iv) <u>Roots are 1/α and 1/β, then:</u>
- (x - 1/α)(x - 1/β) = 0
- x² - (1/α+1/β)x + 1/αβ = 0
- x² - ((α+β)/αβ)x + 1/αβ = 0
- x² - (5/4)/(-1/4)x - 1/(-1/4) = 0
- x² + 5x + 4 = 0
(v) <u>Roots are 2/α² and 2/β², then:</u>
- (x - 2/α²)(x - 2/β²) = 0
- x² - (2/α² + 2/β²)x + 4/(αβ)² = 0
- x² - 2((α+β)² - 2αβ)/(αβ)²)x + 4/(αβ)² = 0
- x² - 2((5/4)² - 2(-1/4))/(-1/4)²x + 4/(-1/4)² = 0
- x² - 2(25/16 + 8/16)/(1/16)x + 4(16) = 0
- x² - 2(33)x + 64 = 0
- x² - 66x + 64 = 0
Answer:
25%
Step-by-step explanation:
let the Market Price of the Watch (M.P.) be x
CP for Rahim=x-[(20/100)*x]
CP = (1 *x) /5 = 4x/5
SP for Rahim = MP = x
Difference (Profit) = SP-CP = x - (4x/5) = (5x - 4x)/5 = x/5
Profit % = (Profit/CP) * 100
Profit % = (x/5)/(4x/5) * 100
Profit % = 25 %
Answer:
Option a) Type I error would occur if we reject null hypothesis and conclude that the average amount is greater than $3,200 when in fact the average amount is $3,200 or less.
Step-by-step explanation:
We are given the following information in the question:
where μ is the average amount of money in a savings account for a person aged 30 to 40.
Type I error:
- Type I error is also known as a “false positive” and is the error of rejecting a null hypothesis when it is actually true.
- In other words, this is the error of accepting an alternative hypothesis when the results can be attributed by null hypothesis.
- A type I error occurs during the hypothesis testing process when a null hypothesis is rejected, even though it is correct and should not be rejected.
Thus, in the above hypothesis type error will occur when we reject the null hypothesis even when it is true.
Option a) Type I error would occur if we reject null hypothesis and conclude that the average amount is greater than $3,200 when in fact the average amount is $3,200 or less.
