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2 years ago

13. For the arithmetic sequence: 6, 12, 18, 24,...

Mathematics

1 answer:

serg [7]2 years ago

4 0

a_15 = 90

a_n = 6n

Step-by-step explanation:

Given sequence is:

6, 12, 18, 24,...

First of all we have to find the common difference. The common difference is the difference between consecutive terms of an arithmetic sequence.

Here,

$d=a_2-a_1 = 12-6 = 6\\d=a_3-a_2= 18-12 = 6$

The general form for arithmetic sequence is:

$a_n=a_1+(n-1)d$

Putting the values for a_1 and d

$a_n=6+(n-1)(6)\\a_n=6+6n-6\\a_n=6n$

a) Find a15

$a_{15} = 6(15)\\=90$

b) Write an equation for the nth term.

The equation is: a_n=6n

Keywords: Arithmetic sequence, Common Difference

Learn more about arithmetic sequence at:

#LearnwithBrainly

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Step-by-step explanation:

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2 years ago

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A study conducted at a certain high school shows that 72% of its graduates enroll at a college. Find the probability that among

mixas84 [53]

Answer:

$P(X \geq 1) =1-P(X$

And we can use the probability mass function and we got:

$P(X=0)=(4C0)(0.72)^0 (1-0.72)^{4-0}=0.00615$

And replacing we got:

$P(X \geq 1) = 1-0.00615 = 0.99385$

Step-by-step explanation:

Let X the random variable of interest "number of graduates who enroll in college", on this case we now that:

$X \sim Binom(n=4, p=0.72)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We want to find the following probability:

$P(X \geq 1)$

And we can use the complement rule and we got:

$P(X \geq 1) =1-P(X$

And we can use the probability mass function and we got:

$P(X=0)=(4C0)(0.72)^0 (1-0.72)^{4-0}=0.00615$

And replacing we got:

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2 years ago

The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c

Anestetic [448]

Answer:

a) The mean is $\mu = 60$

b) The standard deviation is $\sigma = 9$

Step-by-step explanation:

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 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.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when $X = 55.5, Z = -0.5$