suppose the people have weights that are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Find the probability that if a person is randomly selected, his weight will be greater than 174 pounds?
Assume that weights of people are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Mean = 177
standard deviation = 26
We find z-score using given mean and standard deviation
z = 
= 
=-0.11538
Probability (z>-0.11538) = 1 - 0.4562 (use normal distribution table)
= 0.5438
P(weight will be greater than 174 lb) = 0.5438
Answer:
V = 5(12)(7) which is F.
Step-by-step explanation:
V = LWH
L = 12
W = 5
H = 7
12 * 5 * 7
CD and DE are equal distance so their equations are equal:
2x+7 = 4(x-3)
2x+7 = 4x-12
Subtract 2x from both sides
7 = 2x-12
Add 12 to both sides
19 = 2x
9.5 = x
Now that u have x just plug it into the equation for DE
4(9.5-3)
38-12
26 is the answer
Two equilateral triangles, but one has sides measuring 3 cm, while the other measures 6 cm. The figures are similar, but not congruent
Answer:
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Explanation:
Simulate (build a table) the growing of the number of pennies for some nights to figure out the pattern:
First night: 1 penny = 2⁰
Second night: 1 × 2 pennies = 2¹
Third night: 2 × 2 = 2²
Fourth nigth: 2² × 2 = 2³
nth night: 2ⁿ⁻¹
You want 2ⁿ⁻¹ ≥ 2,000,000,000
Which you solve in this way:
- n-1 log (2) ≥ log (2,000,000,000)
- n - 1 ≥ log (2,000,000,000) / log (2)
Since n is number of days, it is an integer number, so n ≥ 32.
Hence, she will have a total of more than $ 2 billion after 32 days.
You can prove that by calculating 2³² = 2,147,483,648.
