Simple..
320-304
= 116
Your answer is 116
Happy Studying ^=^
Answer:
d. Approximate the standard normal distribution with the Student's t distribution
(0.2199 ; 0.2327)
Step-by-step explanation:
Given that :
Sample size, n = 31
Sample mean, xbar = 0.2258
Sample standard deviation, s = 0.0188
Confidence interval (C. I) :
xbar ± margin of error
Margin of Error : Tcritical * s/sqrt(n)
Degree of freedom, df = n - 1 = 31 - 1 = 30
Tcritical value :
T0.05/2, 30 = 2.042
Margin of Error = 2.042 * 0.0188/sqrt(31)
Margin of Error = 0.0068949
C. I = 0.2258 ± 0.0068949
Lower boundary : (0.2258 - 0.006895) = 0.2189
Upper boundary : (0.2258 - 0.006895) = 0.2327
(0.2199 ; 0.2327)
For (11, 1): (11 + 3)1 = 14(1) = 14
For (5, 2): (5 + 3)2 = 8(2) = 16 ≠ 14
For (7, 2): (7 + 3)2 = 10(2) = 20 ≠ 14
For (3, 2): (3 + 3)2 = 6(2) = 12 ≠ 14
Therefore, (11, 1) is a solution to this equation.
Answer:
12 + 8k² + k³ - 9k⁴
Step-by-step explanation:
(8 + k³ - 6k⁴) + (4 - 3k⁴ + 8k²)
8 + k³ - 6k⁴ + 4 - 3k⁴ + 8k²
(8 + 4) + (8k²) + (k³) + (-6k⁴ -3k⁴)
(12) + (8k²) + (k³) + (-9k⁴)
12 + 8k² + k³ - 9k⁴
