Answer:
0.75 mg
Step-by-step explanation:
From the question given above the following data were obtained:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Amount remaining (N) =.?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Number of half-lives (n) =?
n = t / t₁/₂
n = 6/6
n = 1
Finally, we shall determine the amount of the sample remaining after 6 years (i.e 1 half-life) as follow:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Number of half-lives (n) = 1
Amount remaining (N) =.?
N = 1/2ⁿ × N₀
N = 1/2¹ × 1.5
N = 1/2 × 1.5
N = 0.5 × 1.5
N = 0.75 mg
Thus, 0.75 mg of the sample is remaining.
Answer:
B
Step-by-step explanation:
Here’s what B has:
Min: 24
Q1: 27
Median: 31
Q3: 33
Max: 34
Although Choice A and B were really close with their five number summaries, out of the two of them, only B had a maximum value of 34, like the one in the box plot shown above.
Answer:
Step-by-step explanation:
x² = 19x + 12
⇔ x² - 19x - 12 = 0
Calculating the discriminant :
b² - 4ac = (-19)² - 4×1×(-12) = 409
The discriminant is positive ,then the equation has two solutions.
Answer:
21.759
Step-by-step explanation:
Given that :
Mean (m) = 25
Standard deviation (s) = 12.5
Sample size (n) = 40
α = 90%
The confidence interval is obtained using the relation:
Mean ± Zcritical * s/sqrt(n)
Zcritical at 90% confidence interval = 1.64
25 ± 1.64 * (12.5/sqrt(40))
Lower boundary : 25 - 1.64(1.9764235) = 21.75866546
Upper boundary : 25 + 1.64(1.9764235) = 28.24133454
(21.759, 28.241)
Hence, lower bound of confidence interval is : 21.759
