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2 years ago

Find the 9th term of the geometric sequence -8x^5,-32x^6,-128x^7...

Mathematics

1 answer:

zysi [14]2 years ago

7 0

Answer:

-524288^13

Step-by-step explanation:

4: - 512^8

5: -2048^9

6: - 8192^10

7: - 32768^11

8: - 131072^12

9: - -524288^13

Formula:

-n×4^1,2,3......

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2 years ago

The heights of Karina, Miya and Clint were 155cm, 158cm, and 161cm respectively. What is the height of Roger if the mean of all

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3 years ago

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Answer the picture below.

inysia [295]

Answer:

Step-by-step explanation:

function: all the points

note: x intercept is the domain, y intercept is the range

domain must paired with exact range

the domain can be more than 1, but range must be 1.

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3 years ago

Lotteries are an important income source for various governments around the world. However, the availability of lotteries and ot

SashulF [63]

Answer: 0.0473

Step-by-step explanation:

Given : The proportion of gambling addicts : p=0.30

Let x be the binomial variable that represents the number of persons are gambling addicts.

with parameter p=0.30 , n= 10

Using Binomial formula ,

$P(x)=^nC_xp^x(1-p)^{n-x}$

The required probability = $P(x>5)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\=^{10}C_6(0.3)^6(0.7)^{4}+^{10}C_7(0.3)^7(0.7)^{3}+^{10}C_8(0.3)^8(0.7)^{2}+^{10}C_9(0.3)^9(0.7)^{1}+^{10}C_{10}(0.3)^{10}(0.7)^{0}\\\\=\dfrac{10!}{4!6!}(0.3)^6(0.7)^{4}+\dfrac{10!}{3!7!}(0.3)^7(0.7)^{3}+\dfrac{10!}{2!8!}(0.3)^8(0.7)^{2}+\dfrac{10!}{9!1!}(0.3)^9(0.7)^{1}+(1)(0.3)^{10}\\\\=0.0473489874\approx0.0473$

Hence, the required probability = 0.0473

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3 years ago

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DENIUS [597]

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2 years ago