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diamong [38]
3 years ago
5

How to find missing terms in a arithmetic sequence if there is no middle term?

Mathematics
1 answer:
kozerog [31]3 years ago
5 0
You would use the formula for the specific term you wish to find;
The formula is:
U_{n} = a + (n - 1)d
a = starting value of the sequence
d = the common difference (i.e. the difference between any two consecutive terms of the sequence)
n = the value corresponding to the position of the desired term in the sequence (i.e. 1 is the first term, 2 is the second, etc.)
Un = the actual vaue of the the term

For example, if we have the arithmetic sequence:
2, 6, 10, 14, ...
And let's say we want to find the 62nd term;
Then:
a = 2
d = 4
(i.e. 6 - 2 = 4, 10 - 6 = 4, 14 - 10 = 4;
You should always get the same number no matter which two terms you find the difference between so long as they are both consecutive [next to each other], otherwise you are not dealing with an arithmetic sequence)
n = 62

And so:
U_{62} = (2) + (62 - 1)(4) \\
= 2 + 61(4) \\
= 2 + 244 \\
= 246
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BigorU [14]

Answer:

D. y=-0.927x+13.634

Step-by-step explanation:

We graph the points on the graph. The graph is attached.

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Let us now calculate b from the point (1,14)

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b=14.9286

So the equation we get is

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3 years ago
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prisoha [69]
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In a large population of college-educated adults, the mean IQ is 118 with a standard deviation of 20. Suppose 200 adults from th
cricket20 [7]

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Let the IQ of the educated adults be X then;

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To find the probability that the sample mean IQ is greater than 120:

P(X > 120) = 1 - P(X < 120)

Standardize the mean IQ using the sampling formula : Z = (X - μ) / σ/sqrt n

Where;  X = sample mean IQ; μ =population mean IQ; σ = population standard deviation and n = sample size

Therefore, P(X>120) = 1 - P(Z < (120 - 118)/20/sqrt 200)

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The P(Z<1.41) can then be obtained from the Z tables and the value is 0.9207

Thus; P(X< 120) = 1 - 0.9207

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mafiozo [28]
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