As 0.1333 is less than 0.5 the answer is
375
Answer:
13.567%
Step-by-step explanation:
We solve the above question, using z score formula
z = (x-μ)/σ, where
x is the raw score = 23.1 ounces
μ is the population mean = 22.0 ounces
σ is the population standard deviation = 1.0 ounce
More than = Greater than with the sign = >
Hence, for x > 23.1 ounces
z = 23.1 - 22.0/1.0
= 1.1
Probability value from Z-Table:
P(x<23.1) = 0.86433
P(x>23.1) = 1 - P(x<23.1)
P(x>23.1) = 1 - 0.86433
P(x>23.1) = 0.13567
Converting to percentage
= 0.13567 × 100
= 13.567%
Therefore, the percentage of regulation basketballs that weigh more than 23.1 ounces is 13.567%
If you would like to calculate the arithmetic mean, geometric mean, and harmonic mean from the following averages, you can calculate this using the following steps:
averages: 56.4, 59.8, 55.8
the number of values: 3
arithmetic mean:
(56.4 + 59.8 + 55.8) / 3 = 57.33
geometric mean:
(56.4 * 59.8 * 55.8)^(1/3) = 57.31
harmonic mean:
3 / (1/56.4 + 1/59.8 + 1/55.8) = 57.28
Answer:
1 cm = 20 km
so 3.5 cm would equal : 3.5 * 20 = 70 km
hope it helps ;)
Step-by-step explanation:
There are multiple solutions and they will vary.
Hope this helps!
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