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3 years ago

What is the recursive rule for the following sequence: -9, -2, 5, 12, ….

Mathematics

1 answer:

tino4ka555 [31]3 years ago

5 0

Answer:

The answer is option (C)=an-1+7

Step-by-step explanation:

A recursive rule is a formula that in which each term is expressed as a function of its preceding term(s), meaning in order to get to the nth term you have to express it in a form of the term that comes before it. In our case the a(n-1) term

So for the sequence -9, -2, 5, 12

The nth term is any number on the sequence and

-2 is the a(n-1) term for -9

5 is the a(n-1) term for -2

12 is the a(n-1) term for 5

So we need to find out what we have to do to the preceding term to get the next.

To get -2 from -9 we have to add 7 to -9; -9+7=-2

To get 5 from -2 we have to add 7 to -2; -2+7=5

To get 12 from 5 we add 7 to 5; 7+5=12

So the recursive rule would be= a n-1+7

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xxMikexx [17]

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8 0

3 years ago

3(2a + 1) = -5(a + 6)

Lemur [1.5K]

3(2a+1)=-5(a+6)
First expand the brackets.
6a+3=-5a-30
Now add 5a to both sides.
11a+3=-30
Then subtract 3 from both sides.
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3 years ago

An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0

zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .