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mckinley2006
28.08.2019 •
Mathematics
Atravel agency did a survey and found that the average local family spends $1900 on a summer vacation. the distribution is normally distributed with standard deviation $390. a. what percent of the families took vacations that cost under $1500. round to the nearest percent.
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Ответ:
The percentage is 15%
Step-by-step explanation:
* Lets revise how to find the z-score
- The rule the z-score is z = (x - μ)/σ , where
# x is the score
# μ is the mean
# σ is the standard deviation
* Lets solve the problem
- The average local family spends $1900 on a summer vacation
- The distribution is normally distributed with standard deviation $390
- We need to find what percent of the families took vacations that
cost under $1500
∵ The mean is $ 1900
∴ μ = 1900
∵ The standard deviation is $390
∴ σ = 390
∵ The vacation cost is under $1500
∴ x = 1500
∵ z-score = (x - μ)/σ
- Substitute the values above in the rule
∴ z-score = (1500 - 1900)/390 = -1.0256
- Lets use the normal distribution table of z
∵ P(z < 1500) = 0.1515
∵ P(x < 1500) = P(z < 1500)
∴ P(x < 1500) = 0.1515
∴ The percentage of the families took vacations that cost under
$1500 is ⇒ 0.15 × 100% = 15%
∴ The percentage is 15%
Ответ:
The balanced equation is
4Fe(s) + 3O2(g) ---> 2Fe2O3(s)
here four moles of Fe will react with O2 to give 2 moles of Fe2O3
mass of Fe reacted = 33.4 grams
moles of Fe reacted = mass of Fe / atomic mass of Fe
moles of Fe reacted = 33.4 / 55.8 = 0.599 moles
moles of Fe2O3 formed = 0.5 X 0.599 = 0.2995 moles
so mass of Fe2O3 formed = moles X molar mass = 0.2995 X 159.69 = 47.83 g