Answer:
Probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is 0.1515.
Step-by-step explanation:
We are given that the amount a Niffler can hold in their pouch is approximately normally distributed with a mean of 25 pounds of shiny objects and a standard deviation of 6.8 pounds.
Let X = <u><em>amount a Niffler can hold in their pouch</em></u>
So, X ~ Normal(
)
The z score probability distribution for normal distribution is given by;
Z = 
~ N(0,1)
where, 
= population mean = 25 pounds
= standard deviation = 6.8 pounds
Now, the probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is given by = P(X > 32 pounds)
P(X > 32 pounds) = P( > 
) = P(Z > 1.03) = 1 - P(Z 
1.03)
= 1 - 0.8485 = 0.1515
<em>The above probability is calculated by looking at the value of x = 1.03 in the z table which has an area of 0.8485.</em>
<em />
Hence, the probability that a Niffler can hold more than 32 pounds of shiny objects in their pouch is 0.1515.
Answer:
B
Step-by-step explanation:
The recommended sample size n for point estimates is:
n = NX / (N + X - 1)
where N is the population size, and X is defined as:
X = Z² p (1 - p) / E²
where Z is the critical value, p is the sample proportion, and E is the margin of error.
Assume a confidence level of level of 95% and a margin of error of 5%.
α = 0.05, Z(α/2) = 1.96
E = 0.05
Of the 40 units tested, 2 had lifespans less than 26 days. So the proportion is:
p = 2/40 = 0.05
Therefore:
X = (1.96)² (0.05) (1 - 0.05) / (0.05)²
X = 73
Given N = 2000:
n = (2000) (73) / (2000 + 73 - 1)
n = 70.45
Rounding, the recommended sample size is 70.
Answer: 187.5
Step-by-step explanation: Divide 75 by 12 to get 6.25 then multiply that by 30 to get 187.5.
M∠3 = m∠5 = 47° (vertical angles)
m∠1 = 90°
m∠2 + ∠m3 = 90° ⇒ m∠2 = 90 - m∠3 = 90 - 47 = 43°
m∠4 + m∠5 = 180° ⇒ m∠4 = 180 - m∠5 = 180 - 47 = 133°
Answer:
m∠1 = 90°
m∠2 = 43°
m∠3 = 47°
m∠4 = 133°
