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Finger [1]
3 years ago
8

Luke has 1/3 of a package of dried apricots. He divides the dried apricots equally into 3 small bags. Luke gives one of the bags

to a friend and keeps the other two bags for himself. What fraction of the oringial package of dried apricots did Luke keep for himself?
Mathematics
2 answers:
Ksju [112]3 years ago
4 0
Answer: 2/9


Explanation:

\cfrac{1}{3} \div 3 =  \cfrac{1}{3} \times  \cfrac{1}{3} = \cfrac{1}{9}

\cfrac{1}{3} -  \cfrac{1}{9} =  \cfrac{3}{9} -  \cfrac{1}{9} =  \cfrac{2}{9}
Nitella [24]3 years ago
3 0
1/3 ÷ 3 = \frac{1}{9} for each

Luke will stay with a total of:

\frac{1}{9} (x 3) = \frac{3}{9}

\frac{3}{9} -  \frac{1}{9} =  \frac{2}{9}

Hope it helped!
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The average human gestation period is 270 days with a standard deviation of 9 days. The period is normally distributed. What is
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Answer:

Probability that a randomly selected woman's gestation period will be between 261 and 279 days is 0.68.

Step-by-step explanation:

We are given that the average human gestation period is 270 days with a standard deviation of 9 days. The period is normally distributed.

Firstly, Let X = women's gestation period

The z score probability distribution for is given by;

         Z = \frac{ X - \mu}{\sigma} ~ N(0,1)

where, \mu = average gestation period = 270 days

            \sigma = standard deviation = 9 days

Probability that a randomly selected woman's gestation period will be between 261 and 279 days is given by = P(261 < X < 279) = P(X < 279) - P(X \leq 261)

         P(X < 279) = P( \frac{ X - \mu}{\sigma} < \frac{279-270}{9} ) = P(Z < 1) = 0.84134

         P(X \leq 261) = P( \frac{ X - \mu}{\sigma} \leq \frac{261-270}{9} ) = P(Z \leq -1) = 1 - P(Z < 1)

                                                           = 1 - 0.84134 = 0.15866

<em>Therefore, P(261 < X < 279) = 0.84134 - 0.15866 = 0.68</em>

Hence, probability that a randomly selected woman's gestation period will be between 261 and 279 days is 0.68.

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