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2 years ago

What does the f(n - 1) represent?

Mathematics

1 answer:

leonid [27]2 years ago

3 0

I'm assuming that you are talking about sequences <3

In the context of a recursive formula where we have "n-1" in sub-index of "a", you can think of "a" as the previous term in the sequence. In the context of an explicit formula like "-5+2(n-1)" "n-1" represents how many times we need to add 2 to the first term to get the nth term.

Dw if you don't understand, sequences and functions isn't an easy topic

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Ms. Lee, a single adult, earned $65,000 in taxable income last year. Use the rate table below to calculate how much she must pay

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3 years ago

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2 years ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

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3 years ago

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2 years ago

Read 2 more answers

Mileage tests are conducted for a particular model of automobile. If a 98% confidence interval with a margin of error of 1 mile

Llana [10]

Answer: 37

Step-by-step explanation:

Given : Significance level : $\alpha:1-0.98=0.02$

Critical value : $z_{\alpha/2}=z_{0.01}=2.33$

margin of error : E= 1 mile per gallon

Population standard deviation: $\sigma=2.6$ miles per gallon.

We know that when the population standard deviation is known then the formula to find the sample size is given by :

$n=(\dfrac{z_{\alpha/2}\times \sigma}{E})^2$

$n=(\dfrac{(2.33)\times 2.6}{1})^2=36.699364\approx37$ [Round to the next whole number.]

Hence, the required number of automobiles should be used in the test = 37

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3 years ago