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3 years ago

Element X is a radioactive isotope such that every 25 years, its mass decreases by

Mathematics

1 answer:

luda_lava [24]3 years ago

6 0

Answer: 26

Step-by-step explanation:

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Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15. N an 1 4 2 −12 3 36 t

Rzqust [24]

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is $\sum_{n=4}^{15} 5(-2)^{n-1}$$$

<h3>What is sequence ?</h3>

Sequence is collection of numbers with some pattern .

Given sequence

$a_{1}=5\\\\a_{2}=-10\\\\\\a_{3}=20$

We can see that

$\frac{a_1}{a_2}=\frac{-10}{5}=-2\\$

and

$\frac{a_2}{a_3}=\frac{20}{-10}=-2\\$

Hence we can say that given sequence is Geometric progression whose first term is 5 and common ratio is -2

Now $n^{th}$ term of this Geometric progression can be written as

$T_{n}= 5\times(-2)^{n-1}$

So summation of 15 terms can be written as

$\sum_{n=4}^{15} T_{n}\\\\$\\$\sum_{n=4}^{15} 5(-2)^{n-1}$$$

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is $\sum_{n=4}^{15} 5(-2)^{n-1}$$$

To learn more about Geometric progression visit : brainly.com/question/14320920

8 0

3 years ago

Which set of data does not contain any outliers?

dolphi86 [110]

Answer:

im not sure, i will try to figure it out for you

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A regulation basketball has a diameter of about 9 in. What is the best approximation for the volume of this ball? 3050 in³ 382 i

Mandarinka [93]

The best apporximation for the volume of this ball is 382 cubic inches.

<h3>What is the volume of the ball?</h3>

The volume of the ball is calculated by the formula V = 4 / 3πr³. This formula is the same in the whole world.

Given

A regulation basketball has a diameter of about 9 in.

The radius of the ball is;

$\rm Radius = \dfrac{Diameter}{2}\\\\Radius=\dfrac{9}{2}\\\\Radius=4.5\ inches$

The volume of the ball is;

$\rm Volume =\dfrac{4}{3}\pi r^3\\\\Volume=\dfrac{4}{3} \times 3.14 \times 4.5^3\\\\Volume =\dfrac{144.53}{3}\\\\Volume = 381.51\\\\Volume = 382 \ In^3 \ apporx$