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3 years ago

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It is known that when a specific type of radish is grown in a certain manner without fertilizer the weights of the radishes prod

uced are normally distributed with a mean of 40g and a standard deviation of 10g. When the same type of radish is grown in the same way except for the inclusion of fertilizer, it is known that the weights of the radishes produced are normally distributed with a mean of 140g and a standard deviation of 40g. Determine the proportion of radishes grown

1 answer:

Marianna [84]3 years ago

8 0

Answer: hello your question is incomplete below is the missing part

Determine the proportion of radishes grown without fertilizers with weight less than 50grams

answer : p( x < 50 ) = 0.84

Step-by-step explanation:

Radish grown in without fertilizers ( normally distributed )

mean = 40 g

standard deviation = 10 g

Radish grown with fertilizers ( normally distributed )

Mean = 140 g

standard deviation = 40 g

<u>Determine the proportion of radishes grown without fertilizers with weight less than 50 grams </u>

lets assume x = radish without fertilizers

y = radish with fertilizers

hence

p( x < 50 ) = P ( (x-μ / б) < $\frac{50-40}{10}$ )

= p ( z < 1 ) = 0.5 + p ( 0 < z < 1 )

p( x < 50 ) = 0.5 + 0.3413 = 0.8413

attached below is the distribution related to this solution

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Step-by-step explanation:

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3 years ago

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Using function concepts, it is found that:

a) The y-intercept is y = 2.5.

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e) The graph is given at the end of the answer.

------------------------------------

The given function is:

$g(x) = 3 - 8\left(\frac{1}{4}\right)^{2-x}$

------------------------------------

Question a:

The y-intercept is g(0), thus:

$g(0) = 3 - 8\left(\frac{1}{4}\right)^{2-0} = 3 - 8\left(\frac{1}{4}\right)^{2} = 3 - \frac{8}{16} = 3 - 0.5 = 2.5$

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------------------------------------

Question b:

The horizontal asymptote is the limit of the function when x goes to infinity, if it exists.

$\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2+\infty} = 3 - 8\left(\frac{1}{4}\right)^{\infty} = 3 - 8\frac{1^{\infty}}{4^{\infty}} = 3 -0 = 3$

--------------------------------------------------

$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2-\infty} = 3 - 8\left(\frac{1}{4}\right)^{-\infty} = 3 - 8\times 4^{\infty} = 3 - \infty = -\infty$

Thus, the horizontal asymptote is x = 3.

--------------------------------------------------

Question c:

The limit of x going to infinity of the function is negative infinity, which means that the function is decreasing.

--------------------------------------------------

Question d:

Exponential function has no restrictions in the domain, so it is all real values, that is $(-\infty,\infty)$ .
From the limits in item c, the range is: $(-\infty,3)$

--------------------------------------------------

The sketching of the graph is given appended at the end of this answer.

A similar problem is given at brainly.com/question/16533631

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2 years ago

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2 years ago

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3 years ago