Let <em>X</em> be the random variable representing the amount (in grams) of nicotine contained in a randomly chosen cigarette.
P(<em>X</em> ≤ 0.37) = P((<em>X</em> - 0.954)/0.292 ≤ (0.37 - 0.954)/0.292) = P(<em>Z</em> ≤ -2)
where <em>Z</em> follows the standard normal distribution with mean 0 and standard deviation 1. (We just transform <em>X</em> to <em>Z</em> using the rule <em>Z</em> = (<em>X</em> - mean(<em>X</em>))/sd(<em>X</em>).)
Given the required precision for this probability, you should consult a calculator or appropriate <em>z</em>-score table. You would find that
P(<em>Z</em> ≤ -2) ≈ 0.0228
You can also estimate this probabilty using the empirical or 68-95-99.7 rule, which says that approximately 95% of any normal distribution lies within 2 standard deviations of the mean. This is to say,
P(-2 ≤ <em>Z</em> ≤ 2) ≈ 0.95
which means
P(<em>Z</em> ≤ -2 or <em>Z</em> ≥ 2) ≈ 1 - 0.95 = 0.05
The normal distribution is symmetric, so this means
P(<em>Z</em> ≤ -2) ≈ 1/2 × 0.05 = 0.025
which is indeed pretty close to what we found earlier.
Answer:
96 cm^3
Step-by-step explanation:
I hope this helped!
Let the distance of Judy from intersection is x and distance of Jackie from intersection is y.
We convert the given information to equations, step by step.
Point 1: When Jackie is 1 mile farther from the intersection than Judy.
This means y is 1 mile more than x.
So,
y = 1 + x
Point 2: The distance between them is 2 miles more than Judy’s distance from the intersection.
Distance between is x+ y.
So, x+y is 2 miles more than y.
x+y = y + 2
⇒
x = 2
From point 1 we have:
y = 1 + x = 1+ 2 = 3
So,
Distance of Judy from intersection is 2 miles and distance of Jackie from intersection is 3 miles.
Answer:
124
Step-by-step explanation:
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i hope this helps :)
Answer:
1 solution
Step-by-step explanation:
