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lisabon 2012 [21]
3 years ago
12

A picture framer has a thin board 10 1/2 feet long. The framer notices that 2 3/8 feet of the board is scratched and cannot be u

sed. The rest of the board will be used to make small picture frames. Each picture frame needs 1 2/3 feet of the board. At most, how many complete picture frames can be made?
Mathematics
1 answer:
Anestetic [448]3 years ago
6 0

1. Find how many of the board can be used for picture frames:

10\dfrac{1}{2}-2\dfrac{3}{8}=\dfrac{21}{2}-\dfrac{19}{8}=\dfrac{21\cdot 4-19}{8}=\dfrac{65}{8}\ ft.

2. Each picture frame needs 1\dfrac{2}{3}=\dfrac{5}{3} feet of the board, then the farmer can complete

\dfrac{65}{8}:\dfrac{5}{3}=\dfrac{65}{8}\cdot \dfrac{3}{5}=\dfrac{39}{8}=4\dfrac{7}{8}

picture frames.

Answer: at most he can complete 4 picture frames.

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The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de
blsea [12.9K]

Answer:

1. 15.87%

2.  6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

  • Almost all the data lies within three standard deviation from the mean for a normally distributed data.
  • About 68% of data lies within one standard deviation from the mean.
  • About 95% of data lies within two standard deviations of the mean.
  • About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)

Calculation the value from standard normal z table, we have,  

P(x < 6) =0.0228 = 2.28\%

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%

84% of newborn babies weigh between 6.7 and 9.5 pounds.

7 0
3 years ago
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