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Alja [10]
2 years ago
9

Apple weights in an orchard are normally distributed. From a sample farmer Fred determines the mean weight of a box of apples to

be 270 oz. with a standard deviation of 10 oz. He wonders what percent of the apple boxes he has grouped for sale will have a weight less than 255 oz.

Mathematics
2 answers:
Sonja [21]2 years ago
8 0

Answer:

2.07 is the answer as of e.2020

Step-by-step explanation:

ollegr [7]2 years ago
5 0

Answer:

2.07 is the answer as of e.2020

Step-by-step explanation:

Mean = 270 Standard deviation = 10 x = 255 Formula for z-score, z = (x - mean)/SD z = (255 - 270) / 10 => z = -15 / 10 => z = -1.5 So by referring to z-table, -1.5 correlates to 0.0668 that implies to 0.07 So 7% of the boxes of Apples weight less than 255oz. The percentage of boxes is in the range of 255 oz and 270 oz, Now calculating the requiring percentage 50% - 7% = 43%

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Jenna uses 6.1 pints of white paint and blue paint to paint her bedroom walls. 4/5 of this amount is white paint, and the rest i
Novay_Z [31]

Answer:

2.44 pints of white

Step-by-step explanation:

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2 years ago
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soldier1979 [14.2K]

Answer: 8%

Step-by-step explanation:

the answer is 7.5% but you said its a whole number so i guess its 8%

8 0
2 years ago
Bath and 1 yard of ribbon. She is 1/2 yard for project. She wants to divide the rest of the ribbon into pieces 1/4 yards long. H
ikadub [295]

Answer:

2

Step-by-step explanation:

Given:

Total length of ribbon = 1 yard

Length of ribbon used for project = \frac{1}{2} yard

Length of ribbon the rest of ribbon is to be divided = \frac{1}{4} yard

To find:

The number of ribbons of length \frac{1}{4} yard that can be made = ?

Solution:

Length of ribbon left after the 1 yard ribbon is used for project can be calculated by subtracting the length of ribbon used from the initial length of ribbon.

i.e.

Length of ribbon left = 1-\frac{1}{2} =\frac{1}{2} yard

Now, number of ribbon of length \frac{1}{4} yard can be found be dividing the length of ribbon left with the length of ribbon pieces to be cut.

i.e.

Number of ribbons:

\dfrac{\frac{1}{2}}{\frac{1}{4}}\\\Rightarrow \dfrac{1}{2}\times 4 = 2

3 0
3 years ago
Find two positive numbers satisfying the given requirements.
Umnica [9.8K]

Answer:

8 and 19

Step-by-step explanation:

To some this, let's first list all the factors of 152. They are;

1, 2, 4, 8, 19, 38, 76, 152.

Now, let's arrange them to reflect being multiplied to get 152.

Thus;

1 × 152 = 152

2 × 76 = 152

4 × 38 = 152

8 × 19 = 152

Also, let's do the same for their sum;

1 + 152 = 153

2 + 76 = 78

4 + 38 = 42

8 + 19 = 27

Looking at the figures above, the ones that their product is 152 but have the least sum are 8 and 19

8 0
2 years ago
It is believed that 36% of the US population has never been married, 32% are divorced, 27% are married, and 5% are widowed. You
Katena32 [7]

Answer: the smallest number of people required for the sample to meet the conditions for performing inference is 100

Step-by-step explanation:

Given that;

36% of US population has never been married

32% are divorced

27% are married

5% are widowed

Taking a simple random sample of individuals to test this claim;

we need expected count in each cell to be at least 5, here the smallest proportion is 5% = 0.05

so we only need to satisfy condition for its expected count;

n × 0.05 ≥ 5

n = 5 / 0.05 = 100

Therefore the smallest number of people required for the sample to meet the conditions for performing inference is 100

6 0
2 years ago
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