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4 years ago

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A grocery store purchases melons from two distributors, J & K. Distributor J provides melons from organic farms. The distrib

ution of the diameters of the melons from distributor J is approximately normal with mean with 133 millimeters and standard deviation 5 mm. For a melon selected at random from distributor J, what is the probability that the melon will have a diameter greater than 137mm?

2 answers:

harina [27]4 years ago

6 0

Answer: green.

Step-by-step explanation:

Vsevolod [243]4 years ago

3 0

Answer: 0.2119

Step-by-step explanation:

Find the z-score.

(x - μ)/σ

The equation would be:

(137-133)/5 = 0.8

p-value = 0.7881

Subtract that from 1.

1 - 0.7881 = 0.2119

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Consider a sample with data values of 27, 24, 21, 16, 30, 33, 28, and 24. Compute the 20th, 25th, 65th, and 75th percentiles. 20

densk [106]

Answer:

$P_{20} = 20$ --- 20th percentile

$P_{25} = 21.75$ --- 25th percentile

$P_{65} = 27.85$ --- 65th percentile

$P_{75} = 29.5$ --- 75th percentile

Step-by-step explanation:

Given

27, 24, 21, 16, 30, 33, 28, and 24.

N = 8

First, arrange the data in ascending order:

Arranged data: 16, 21, 24, 24, 27, 28, 30, 33

Solving (a): The 20th percentile

This is calculated as:

$P_{20} = 20 * \frac{N +1}{100}$

$P_{20} = 20 * \frac{8 +1}{100}$

$P_{20} = 20 * \frac{9}{100}$

$P_{20} = \frac{20 * 9}{100}$

$P_{20} = \frac{180}{100}$

$P_{20} = 1.8th\ item$

This is then calculated as:

$P_{20} = 1st\ Item +0.8(2nd\ Item - 1st\ Item)$

$P_{20} = 16 + 0.8*(21 - 16)$

$P_{20} = 16 + 0.8*5$

$P_{20} = 16 + 4$

$P_{20} = 20$

Solving (b): The 25th percentile

This is calculated as:

$P_{25} = 25 * \frac{N +1}{100}$

$P_{25} = 25 * \frac{8 +1}{100}$

$P_{25} = 25 * \frac{9}{100}$

$P_{25} = \frac{25 * 9}{100}$

$P_{25} = \frac{225}{100}$

$P_{25} = 2.25\ th$

This is then calculated as:

$P_{25} = 2nd\ item + 0.25(3rd\ item-2nd\ item)$

$P_{25} = 21 + 0.25(24-21)$

$P_{25} = 21 + 0.25(3)$

$P_{25} = 21 + 0.75$

$P_{25} = 21.75$

Solving (c): The 65th percentile

This is calculated as:

$P_{65} = 65 * \frac{N +1}{100}$

$P_{65} = 65 * \frac{8 +1}{100}$

$P_{65} = 65 * \frac{9}{100}$

$P_{65} = \frac{65 * 9}{100}$

$P_{65} = \frac{585}{100}$

$P_{65} = 5.85\th$

This is then calculated as:

$P_{65} = 5th + 0.85(6th - 5th)$

$P_{65} = 27 + 0.85(28 - 27)$

$P_{65} = 27 + 0.85(1)$

$P_{65} = 27 + 0.85$

$P_{65} = 27.85$

Solving (d): The 75th percentile

This is calculated as:

$P_{75} = 75 * \frac{N +1}{100}$

$P_{75} = 75 * \frac{8 +1}{100}$

$P_{75} = 75 * \frac{9}{100}$

$P_{75} = \frac{75 * 9}{100}$

$P_{75} = \frac{675}{100}$

$P_{75} = 6.75th$

This is then calculated as:

$P_{75} = 6th + 0.75(7th - 6th)$

$P_{75} = 28 + 0.75(30- 28)$

$P_{75} = 28 + 0.75(2)$

$P_{75} = 28 + 1.5$

$P_{75} = 29.5$

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Answer:

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We are not given the velocity of the plane at landing but we are given velocities at two time periods

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Plugging this into the velocity equation we get

$\bold{v_5 = a.5 + v_0 = 5a + v_0}$ (1)

Similarly

$\bold{v_{15} = a.15 + v_0 = 15a + v_0}$ (2)

Subtract (1) from $v_{15} - v_5 = (15a + v_0) - (5a + v_0) = (15a-5a) + (v_0-v_0) = 10a$

Therefore $a = \frac{v_{15} - v_5}{10}$

Given $v_{15}= 50 m/s \textrm{ and } v_5 = 110m/s$

this evaluates to

$a = (50-110)/10 =$ -60 m/s²

The negative value is because the plane is actually decelerating which is a negative number

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Readme [11.4K]

You should try mathway with your question :)

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