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ziro4ka [17]
2 years ago
15

A farmer believes that the yields of his tomato plants have a normal distribution with an average yield of 11 lbs and a standard

deviation of 2.1 lbs. The farmer would like to identify the plants which yield the highest 5% in weight and save them for breeding purposes. What yielded weight separates the top 5% of the crop from the rest? Use the information above and the Z table to find the Z SCORE that you will use in this problem.
Mathematics
1 answer:
AnnyKZ [126]2 years ago
5 0
Given:
mean, μ = 11 lb
Std. deviation, σ = 2.1

Let x =  the weight that separates the top 5% of the crop.
The z-score is
z = (x - μ)/σ 
   = (x - 11)/2.1

From standard z-table for normal distribution,
P(z=1.645) = 0.95
 
Therefore
(x - 11)/2.1 = 1.645
x - 11 = 1.645/2.1 = 0.7833
x = 11.783

Answer:  11.8 lb (nearest tenth)
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<u>Given</u>:

The 11th term in a geometric sequence is 48.

The 12th term in the sequence is 192.

The common ratio is 4.

We need to determine the 10th term of the sequence.

<u>General term:</u>

The general term of the geometric sequence is given by

a_n=a(r)^{n-1}

where a is the first term and r is the common ratio.

The 11th term is given is

a_{11}=a(4)^{11-1}

48=a(4)^{10} ------- (1)

The 12th term is given by

192=a(4)^{11} ------- (2)

<u>Value of a:</u>

The value of a can be determined by solving any one of the two equations.

Hence, let us solve the equation (1) to determine the value of a.

Thus, we have;

48=a(1048576)

Dividing both sides by 1048576, we get;

\frac{3}{65536}=a

Thus, the value of a is \frac{3}{65536}

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The 10th term of the sequence can be determined by substituting the values a and the common ratio r in the general term a_n=a(r)^{n-1}, we get;

a_{10}=\frac{3}{65536}(4)^{10-1}

a_{10}=\frac{3}{65536}(4)^{9}

a_{10}=\frac{3}{65536}(262144)

a_{10}=\frac{786432}{65536}

a_{10}=12

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