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Katarina [22]
3 years ago
5

Find a formula for the nth term in this arithmetic sequence. a1=0, a2=0.5, a3=1, a4=1.5

Mathematics
1 answer:
MrRissso [65]3 years ago
6 0

Answer:

nth term is;

0.5n-0.5

Step-by-step explanation:

As we can see, the first term is 0

The common difference is 0.5-0= 1-0.5 = 1.5-1 = 0.5

Formula for nth term of an arithmetic sequence is;

a + (n-1)d

So we have

0 + (n-1) 0.5

= 0.5n-0.5

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What is f(x) = –4x2 + 24x + 13 written in vertex form? f(x) = –4(x – 6)2 + 23 f(x) = –4(x – 6)2 + 7 f(x) = –4(x – 3)2 + 4 f(x) =
nika2105 [10]
We have that
f(x) = –4x²<span> + 24x + 13
</span>
we know that

The vertex form for a parabola that opens up or down is:

f(x) = a(x - h)^2 + k

in the given equation, <span>a=-4</span><span>, therefore we add zero to the original equation in the form of </span><span>4h</span>²<span>−4h</span>²
f(x) = –4x² + 24x + 4h²−4h² +13
<span>Factor 4 out of the first 3 terms and group them 
</span>f(x) = –4*(x² -6x +h²) +4h² +13
<span>We can find the value of h by setting the middle term equal to -2hx
</span>−2hx=−6x
<span>h=3</span><span> and  </span><span>4h</span>²<span>=<span>36
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4 0
3 years ago
According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e
slamgirl [31]

Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that p = 0.09

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4)

So

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225

P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827

P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082

P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982

Finally

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.982 = 0.0180

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

6 0
2 years ago
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