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allsm [11]
3 years ago
8

A 2-column table with 5 rows. Column 1 is labeled Decimal with entries 0.1 6 repeating, 0.3 repeating, 0.5, x, 0.8 3 repeating.

Column 2 is labeled Fraction with entries one-sixth, two-sixths, three-sixths, four-sixths, five-sixths.
Which best shows how the table can be used to determine the missing value x?

2(two-sixths) equals four-sixths, so 2(0.5) = 1, and x = 1.
4(1-sixth) equals four-sixths, so 4(0.16) = 0.64, and x = 0.64.
Three-sixths plus one-sixth equals four-sixths, so 0.5 + 0.16 = 0.6, and x = 0.6.
Five-sixths plus one-sixth equals four-sixths, so 0.83 + 0.16 = 0.99, and x = 0.99.
Mathematics
1 answer:
Sindrei [870]3 years ago
4 0

Answer:

A 2-column table with 5 rows. Column 1 is labeled Decimal with entries 0.1 6 repeating, 0.3 repeating, 0.5, x, 0.8 3 repeating. Column 2 is labeled Fraction with entries one-sixth, two-sixths, three-sixths, four-sixths, five-sixths.

Which best shows how the table can be used to determine the missing value x?

2(two-sixths) equals four-sixths, so 2(0.5) = 1, and x = 1.

4(1-sixth) equals four-sixths, so 4(0.16) = 0.64, and x = 0.64.

Three-sixths plus one-sixth equals four-sixths, so 0.5 + 0.16 = 0.6, and x = 0.6.

Five-sixths plus one-sixth equals four-sixths, so 0.83 + 0.16 = 0.99, and x = 0.9

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A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

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

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

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