Is 12+5(2+y) and 5y+22 equivalent
1 answer:
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Answer:
Kimberly and Mike had at first $100 each
Step-by-step explanation:
Assume that Kimberly and Mike each have $x
∵ Kimberly has $x
∵ Kimberly spent $50
- Find the reminder with Kimberly by subtracting 50 from x
∴ The money left with Kimberly = x - 50
∵ Mike has $x
∵ Mike spent $25
- Find the reminder with Mike by subtracting 25 from x
∴ The money left with Mike = x - 25
∵ Mike's money is 50% more than Kimberly's money
- That means Mike's money is more than Kimberly's money by
50% of Kimberly's money
∴ x - 25 = (x - 50) + 50%(x - 50)
∵ 50% = = 0.5
∴ x - 25 = (x - 50) + 0.5(x - 50)
- Simplify the right hand side
∴ x - 25 = x - 50 + 0.5x - 25
- Add the like terms in the right hand side
∴ x - 25 = 1.5x - 75
- Add 75 to both sides
∴ x + 50 = 1.5x
- Subtract x from both sides
∴ 50 = 0.5x
- Divide both sides by 0.5
∴ 100 = x
∵ x represents the amount of money that Kimberly and
Mike each had at first
∴ Kimberly and Mike had at first $100 each
The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.
That is,
Consider X be the length of the pregnancy
Mean and standard deviation of the length of the pregnancy.
Mean
Standard deviation \sigma =15
For part (a) , to find the probability of a pregnancy lasting 308 days or longer:
That is, to find
Using normal distribution,
Thus
So
Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.
This the answer for part(a): 0.00256
For part(b), to find the length that separates premature babies from those who are not premature.
Given that the length of pregnancy is in the lowest 3%.
The z-value for the lowest of 3% is -1.8808
Then
This implies
Thus the babies who are born on or before 238 days are considered to be premature.
