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serious [3.7K]
2 years ago
9

Find the first term and the common ratio third term =6 and seventh term= 96​

Mathematics
1 answer:
mamaluj [8]2 years ago
6 0

Answer:

First term a₁ = 3/2 and common ratio r = 2

Step-by-step explanation:

We need to find the first term and common ratio while we are given third term = 6 and seventh term = 96

Since common ratio is required so, the sequence is geometric sequence

The formula used is: a_n= a_1r^{n-1}

We are given: third term = 6 i,e

a_3=a_1r(3-1)\\6=a_1r^2----eq(1)

Seventh term = 96

a_7=a_1r(7-1)\\96=a_1r^6----eq(2)

Dividing eq(2) and eq(3)

\frac{96}{6}=\frac{a_1r^6}{a1r2}\\16=\frac{r^6}{r^2}\\16=r^{6-2}\\=> r^4=16\\Taking \ fourth \ root\\ r=2,-2,2i,-2i\\ \ We \ will \ consider \ only \ positive \ value \ of \ r \ i.e \ \textbf{ r= 2}

So, Common Ratio r = 2

Finding First term using eq(1)

a_3=a_1r^2\\6=a_1(2)^2\\6=a_1(4)\\a_1= \frac{6}{4}\\a_1= \frac{3}{2}

So, First term a₁ = 3/2 and common ratio r = 2

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Answer:

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Step-by-step explanation:

Z-score:

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Idonna scored 671 on the mathematics part of the SAT. The distribution of SAT math scores in that year was Normal with mean 509 and standard deviation 115.

This means that her standardized score is Z when X = 671, \mu = 509, \sigma = 115. So

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Idonna's standardized score is 1.41.

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