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aksik [14]
3 years ago
5

How much smaller is the sum of the first 1000 natural numbers than the sum of the first 1001 natural numbers?

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
5 0

ANSWER

1001

EXPLANATION

The sum of the first n - natural numbers is

S_n=  \frac{n}{2} (2a + d(n - 1))

The sum of the first 1000 terms is

S_{1000}=  \frac{1000}{2} (2(1) + 1(1000 - 1))

S_{1000}=500 (1001)

S_{1000}=500500

The sum of the first 1001 terms is

S_{1001}=  \frac{1001}{2} (2(1) + 1(1001 - 1))

S_{1001}= 1001 \times (501)

= 501501

The difference is

501501 - 500500= 1001

sesenic [268]3 years ago
3 0

Answer:

The answer to this questions is 1001.

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Answer:

a) X        Freq.         Rel Freq.

__________________________

0           7              (7/58) = 0.121

1            10            (10/58)=0.172

2           13            (13/58) =0.224

3           14            (14/58)=0.241

4           6              (6/58)=0.103

5           3              (3/58)=0.052

6           3              (3/58)=0.052

7            1              (1/58) =0.017

8            1              (1/58)=0.017

_________________________

Total     58                 1.00

b) \frac{7+10+13+14+6}{58}= \frac{50}{58}=0.862

Step-by-step explanation:

Assuming this question: Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined, resulting in the following data:

2,1,2,4,0,1,3,2,0,5,3,3,1,3,2,4,7,0,2,3,

0,4,2,1,3,1,1,3,4,2,3,2,2,8,4,5,1,3,1,

5,0,2,3,2,1,0,6,4,2,1,6,0,3,3,3,6,2,3

(a) Determine frequencies and relative frequencies for the observed values of x = number of nonconforming transducers in a batch. (Round your relative frequencies to three decimal places.)

For this case first we order the dataset on increasing way and we got:

0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1

2 2 2 2  2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3 3 3 3  3 3

4 4 4 4 4 4

5 5 5

6 6 6

7

8

And we can construct the following table:

X        Freq.         Rel Freq.

__________________________

0           7              (7/58) = 0.121

1            10            (10/58)=0.172

2           13            (13/58) =0.224

3           14            (14/58)=0.241

4           6              (6/58)=0.103

5           3              (3/58)=0.052

6           3              (3/58)=0.052

7            1              (1/58) =0.017

8            1              (1/58)=0.017

_________________________

Total     58                 1.00

(b) What proportion of batches in the sample have at most four nonconforming transducers? (Round your answer to three decimal places.)

For this case this proportion would be:

\frac{7+10+13+14+6}{58}= \frac{50}{58}=0.862

4 0
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