Compute 1 + 2 + 3 +....+ 1,997 + 1,998 + 1,999

Mathematics

1 answer:

sp2606 [1]2 years ago

7 0

Answer:

1 999 000

Step-by-step explanation:

Formula:

$1+2+3+.\ .\ .+n=\frac{n\times \left( n+1\right) }{2}$

………………………………………

Then

$1+2+3+....+1997+1998+1999=\frac{1999\times \left( 1999+1\right) }{2}$

$1+2+3+....+1997+1998+1999=\frac{1999\times \left( 2000\right) }{2}$

$1+2+3+....+1997+1998+1999=\frac{3998000 }{2}$

$1+2+3+....+1997+1998+1999=1999000$

You might be interested in

The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What

atroni [7]

Answer:

21.77% probability that a randomly selected teacher earns more than $525 a week

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 490, \sigma = 45$