Is (1, 2) a solution of the system?

Use Gauss's approach to find the following sum (do not use formulas):5+11+17+23...+83

LUCKY_DIMON [66]

Denote the sum by S. So

S = 5 + 11 + 17 + 23 + ... + 83

There's a constant difference of 6 between consecutive terms in S, so the 3 terms before 83 are 77, 71, and 65. So

S = 5 + 11 + 17 + 23 + ... + 65 + 71 + 77 + 83

Gauss's approach involves inverting the sum:

S = 83 + 77 + 71 + 65 + ... + 23 + 17 + 11 + 5

If we add terms in the same position in the sums, we get

2S = (5 + 83) + (11 + 77) + ... + (77 + 11) + (83 + 5)

and we notice that each grouped term on the right gives a total of 88. So the right side consists of several copies n of 88, which means

2S = 88n

and dividing both sides by 2 gives

S = 44n

Now it's a matter of determining how many copies get added. The terms in the sum form an arithmetic progression that follows the pattern

11 = 5 + 6

17 = 5 + 2*6

23 = 5 + 3*6

and so on, up to

83 = 5 + 13*6

so n = 13, which means the sum is S = 44*13 = 572.

8 0

3 years ago

If there are 225 pets in the complex, approximately how many cats are there?

zepelin [54]

Answer:

about 80 cats

Step-by-step explanation:

You multiply 225 pets by the 35% of cats giving you the equation of 225×.35 giving you 78.75. But you approximate the 78 which gives you 80. Thus giving the the 35% of cats of 80 cats.

5 0

3 years ago

Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are norma

Pachacha [2.7K]

Answer:

Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.

There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

Step-by-step explanation:

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

$Z = \frac{X - \mu}{\sigma}$

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem

Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that $\mu = 14.4, \sigma = 1.1$ .

What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?

For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16.2 - 14.4}{1.1}$