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3 years ago

Identify the 25th term of the arithmetic sequence 2, 1 and 3 over 5,

Mathematics

2 answers:

Katarina [22]3 years ago

5 0

-7 3/5 is the correct answer simplified.

raketka [301]3 years ago

3 0

First, you have to transform your numbers into proper fractions:
1 and 3/5 = (5+3)/5 = 8/5
1 and 1/5 = (5+1)/5 = 6/5

Therefore your arithmetic sequence is 2, 8/5, 6/5, ...

In an arithmetic sequence, the difference between a given term and the preceding one is equal to the difference between the <span>following </span>term and the given one. In your case:
d = 8/5 - 2 = (8-10)/5 = -2/5
As a prove: d = 6/5 - 8/5 = -2/5

Now, in order to find the 25th term you need to apply the formula:
an = a + (n - 1)d
where an is the number you are looking for, a is the first term, n is the term you are looking for and d is the distance.
Hence,
a₂₅ = 2 + (25-1)(-2/5) = 2 - 24·2/5 = 2 - 48/5 = (10-48)/5 = -38/5

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25 : 750
(÷25) (÷25)
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The sample is 5 when corrected to the nearest whole number.
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3 years ago

dimulka [17.4K]

So car A travels Distance = 60 km/hr*(1+x)

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60 +60x = 75x

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x=4, but my expression is written to count from when car B commenced travel. So total time is 5 hours from the car A setting off.

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telo118 [61]

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2 years ago

According to a recent publication, the mean price of new mobile homes is $63 comma 800. Assume a standard deviation of $7900.

wolverine [178]

Answer:

a. For n=25, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are $63,800 and $1,580, respectively.

b. For n=50, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are $63,800 and $1,117, respectively.

Step-by-step explanation:

In this case, for each sample size, we have a sampling distribution (a distribution for the population of sample means), with the following parameters:

$\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{n}$

For n=25 we have:

$\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{25}=7,900/5=1,580$

The spread of the sampling distribution is always smaller than the population spread of the individuals. The spread is smaller as the sample size increase.

This has the implication that is expected to have more precision in the estimation of the population mean when we use bigger samples than smaller ones.

If n=50, we have:

$\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{50}=7,900/7.07=1,117$

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4 years ago