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Karo-lina-s [1.5K]
3 years ago
9

What is the some of the counting numbers from 1 to 80?

Mathematics
1 answer:
kumpel [21]3 years ago
7 0
There is a formula for adding consecutive numbers:
total = n*(n+1)/2
total = 80*(81)/2
total = 3,240

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The value of is between which two<br> numbers?
Karo-lina-s [1.5K]

Answer:

12.3693168769

Step-by-step explanation:

Do squareroot(153) and it gives you the answer

6 0
2 years ago
Determine if the sequence is arithmetic. If it is, find the common difference.
Iteru [2.4K]

Answer:

Yes, they are adding 100 every time.

Step-by-step explanation:

4 0
2 years ago
You have two parents, four grandparents, eight greatgrandparents, and so forth.
Leviafan [203]
A.
In the first generation we have 2 ancestors.
In the second generation we have 4 ancestors or 2^{2} ancestors.
In the third generation we have 8 ancestors ot 2^{3} ancestors.

We can see that each succesive generation has two times more members. The sum is:
2+4+8+16+...+ 2^{40}

To calculate number of ancestors we can use formula <span>for the sum of a geometric sequence. Geometric sequence is sequence of numbers that differ by a certain factor. This factor is called ratio. Formula is:
</span>S= a_{1} * \frac{1- r^{n} }{1-r}
<span>Where:
S -> sum
a1 -> first member of a sequence
r -> ratio
n -> number of elements

For this question:
a1 = 2
r = 2
n = 40
</span>S= 2* \frac{1- 2^{40} }{1-2} =2 199 023 255 550
<span>
b.
1 generation = 25 years
40 generations = 40 * 25 = 1000 years

c.
Total number of people who have ever lived = </span>10^{10}
Number of ancestors in 40 generations = 2.2*10^{12}

The number of ancestors is greater than total number of people who have ever lived. This means that not all ancestors were distinct and that in each generation both men and women had children with more than one partner.
8 0
3 years ago
This is the same question as last time just not blurry.
AveGali [126]
4. proportional
5. proportional
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8 0
3 years ago
Read 2 more answers
Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1
alexandr1967 [171]

Answer:

a) \simeq 0.3012   b) \simeq 0.0494 c) \simeq 0.2438

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, \frac{1.2}{4}

= 0.3 collisions per  month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

          P(X =x) = \frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}&#10;

                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, \lambda = 0.3 collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

         P(X = 0) = \frac{e^{-0.3}\times {\0.3}^{0}}{0!}&#10;

                      \simeq 0.7408182207 ------------------------------------(2)

Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =0.7408182207^{4}

     \simeq 0.3012  -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

           P(X =1) = \frac{e^{-0.3}\times {\0.3}^{1}}{1!}&#10;

                      \simeq 0.2222454662 ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

                =0.2222454662^{2}

                \simeq 0.0494 -----------------------------------------(5)

1 collision in 6 months period means

                                \frac{1}{6} collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6]  (which is to be estimated)

=\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

\simeq 0.6543894283-------------------------------------------(6)

So,

P(1  collision in 6 month period)

  =  0.6543894283^{6}

   \simeq 0.0785267444 ------------------------------------------------(7)

So,

P(No collision in 6 months period)

  = (P(X =0)^{6}

   \simeq 0.1652988882 ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

\simeq  0.2438 ---------------------------------------------(9)          

7 0
3 years ago
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