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sergeinik [125]
3 years ago
9

A farmer is planting a field with 28 rows of vegetables, and each row requires 6.491 gallons of water per day. How many gallons

of water each day does the farmer need to water the 28 rows of vegetables? If there are 10 plants in each row, find the daily amount of water delivered to each plant.
Mathematics
2 answers:
Mama L [17]3 years ago
7 0

Since each row requires 6.491 gallons of water per day, that means that you will multiply 6.491 by 28, because there are 28 rows of vegetables. The farmer needs 181.748 gallons per day. To find out how much water each plant needs, you will divide 6.491 by 10, because there are 10 plants in each row. The answer should be 0.6491 gallons of water per plant.

Andreyy893 years ago
5 0

28 rows times 6.491 gallons/day = 181.75 gallons/day.


That's per row. Each of the 10 plants in each row would receive



6.491 gallons water per row

------------------------------------------- = 0.649 gallon/plant

10 plants per row

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Margarita [4]

Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 100 and variance of 36.

This means that \mu = 100, \sigma = \sqrt{36} = 6

Sample of 25:

This means that n = 25, s = \frac{6}{\sqrt{25}} = 1.2

(a) P(X<99)

This is the pvalue of Z when X = 99. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{99 - 100}{1.2}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

Z = \frac{X - \mu}{s}

Z = \frac{100 - 100}{1.2}

Z = 0

Z = 0 has a pvalue of 0.5

X = 98

Z = \frac{X - \mu}{s}

Z = \frac{98 - 100}{1.2}

Z = -1.67

Z = -1.67 has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525

6 0
2 years ago
Factor the polynomial.<br> 13x² – 20x – 12
erma4kov [3.2K]

Answer:

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Step-by-step explanation:

Factor the following:

13 x^2 - 20 x - 12

Factor the quadratic 13 x^2 - 20 x - 12. The coefficient of x^2 is 13 and the constant term is -12. The product of 13 and -12 is -156. The factors of -156 which sum to -20 are 6 and -26. So 13 x^2 - 20 x - 12 = 13 x^2 - 26 x + 6 x - 12 = x (13 x + 6) - 2 (13 x + 6):

x (13 x + 6) - 2 (13 x + 6)

Factor 13 x + 6 from x (13 x + 6) - 2 (13 x + 6):

Answer: (13 x + 6) (x - 2)

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Step-by-step explanation:

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Answer:

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