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tekilochka [14]
3 years ago
10

2,5,8,11,14 write down in terms of n , an expression for the nth term for this sequence

Mathematics
1 answer:
Alisiya [41]3 years ago
4 0

Answer:

a_{n} = 3n - 1

Step-by-step explanation:

Note there is a common difference d between consecutive terms in the sequence, that is

5 - 2 = 8 - 5 = 11 - 8 = 14 - 11 = 3

This indicates the sequence is arithmetic with n th term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 2 and d = 3, thus

a_{n} = 2 + 3(n - 1) = 2 + 3n - 3 = 3n - 1

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Some students are going on a field trip. The cost of the trip is $100, in addition to $7per student. Write an equation that rela
grigory [225]

Answer: C= 100+7n

Step-by-step explanation:

5 0
2 years ago
Which results only in a horizontal compression of y= 1/х by a factor of 6?
marta [7]

Answer:

I'm assuming you mean a compression of factor 6

In that case, it will be h = 1/6x

5 0
3 years ago
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
3 years ago
Can some one explain how to do this?
Delvig [45]

Answer:

see explanation

Step-by-step explanation:

Using the tangent and sine ratios in the right triangle EFG

tan60° = \frac{opposite}{adjacent} = \frac{FG}{EG} = \frac{28}{EG} ( multiply both sides by EG )

EG × tan60° = 28 ( divide both sides by tan60° )

EG = \frac{28}{tan60} ≈ 16.2 in ( to the nearest tenth )

--------------------------------------------------------------

sin60° = \frac{opposite}{hypotenuse} = \frac{FG}{EF} = \frac{28}{EF} ( multiply both sides by EF )

EF × sin60° = 28 ( divide both sides by sin60° )

EF = \frac{28}{sin60} ≈ 32.3 in ( to the nearest tenth )

3 0
3 years ago
11,20,29, Find the 30th term.
Mekhanik [1.2K]

Answer:

372

There are already 3 terms. So, now you have to find the 30th term.

The difference of 30 - 3 is 27. Now, since you add 9 for every term, multiply it by 27.

27 x 9 = 243.

The third term is 29. So you add 29.

Your answer will be 372. Hope it helps.

4 0
3 years ago
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