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3 years ago

Determine the distance between -10 and 5. HELP

Mathematics

2 answers:

kow [346]3 years ago

7 0

Answer: 15

Step-by-step explanation: Good luck! :D

OLga [1]3 years ago

6 0

15, I hope this helps

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Please please please help!! i’ll mark the brainliest and give 15 points!!!!

Korolek [52]

9514 1404 393

Answer:

-108

Step-by-step explanation:

About the easiest way to do this for small values of n is to compute each of the terms using the given recurrence relation.

$a_1=4\\\\a_2=-3a_1=-3(4)=-12\\\\a_3=-3a_2=-3(-12)=36\\\\a_4=-3a_3=-3(36)=-108\\\\\boxed{a_4=-108}$

_____

Alternate solution

You recognize that the recurrence relation describes a geometric sequence with a first term of 4 and a common ratio of -3. The n-th term of a geometric sequence is ...

$a_n=a_1\cdot r^{n-1} \qquad\text{for first term $a_1$ and common ratio $r$}$

Then the 4th term will be ...

$a_4=4\cdot(-3)^{4-1}=4\cdot(-27)=-108$

3 0

3 years ago

Combine like terms 4y^2+4(7y^2-8)= please show your work

kirza4 [7]

Answer:

32y^2-32

Step-by-step explanation:

4y^2+4(7y^2-8)

4y^2+28y^2-32

32y^2-32

if you need extra steeps explaining then plz let me know and I will help :)

8 0

3 years ago

SOMEONE HELP ME WITH THIS ASAP FOR BRAINLIEST THANK YOU!

antiseptic1488 [7]

Answer:1. (1,2) 2.(-1,0)

Step-by-step explanation:

1. 5x-4(-3x+5)=3 x=1 2. 3(2y-1)-2y=-3 y=0

y=-3x1+5 y=2 Plug them in X=2x0-1 x=-1 plug them in

4 0

2 years ago

Suppose that minor errors occur on a computer in a space station, which will require re-calculation. Assume the occurrence of er

Molodets [167]

Answer:

$P(X = 0) = 0.6065$

$P(x < 25 ) = 1.18 *10^{-33}$

$P(x \le 5 ) = 0.9994$

Step-by-step explanation:

From the question we are told that

The rate is $\lambda = \frac{1}{2}\ hr^{-1}$ = 0.5 / hr

Generally Poisson distribution formula is mathematically represented as

$P(X = x) = \frac{(\lambda t) ^x e^{-\lambda t }}{x!}$

Generally the probability that no error occurred during a day is mathematically represented as

Here t = 1 hour according to question a

$P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$

Hence

$[tex]P(X = 0) = \frac{\frac{1}{2} ^0 e^{-\frac{1}{2}}}{0!}$

=> $P(X = 0) = 0.6065$

Generally the probability that a critical error occurs since the start of a day is mathematically represented as

Here t = 1 hour according to question a

$P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$

Hence

$P(x \ge 25 ) = 1 - P(x < 25 )$

Here

$P(x < 25 ) = \sum_{x=0}^{24} \frac{e^{-\lambda} * \lambda^{x}}{x!}$

=> $P(x < 25 ) = \frac{e^{-0.5} *0.5^{0}}{0!} + \cdots + \frac{e^{-0.5} *0.5^{24}}{24!}$

$P(x < 25 ) = 0.6065 + \cdots + \frac{e^{-0.5} *0.5^{24}}{6.204484 * 10^{23}}$

$P(x < 25 ) = 0.6065 + \cdots + 6.0*10^{-32}$

$P(x < 25 ) = 1.18 *10^{-33}$

Considering question c

Here t = 2

Gnerally given that the system just started up and an error occurred the probability the next reset will occur within 2 hours

$P(x \le 5 ) = \sum_{n=0}^{5} \frac{(\lambda t) ^x e^{-\lambda t }}{x!}$

=> $P(x \le 5 ) = \frac{(0.5 * 2) ^ 0 e^{- 0.5 * 2 }}{0!} + \cdots + \frac{(0.5 * 2) ^ 5 e^{- 0.5 * 2 }}{5!}$

=> $P(x \le 5 ) = \frac{1* 2.7183 }{1 } + \cdots + \frac{1 *2.7183 }{120}$

=> $P(x \le 5 ) = 2.7183 + \cdots + 0.0226525$

$P(x \le 5 ) = 0.9994$

8 0

3 years ago

HEEEELLLLLLPPP !!!!!!!! What is the product ?

Roman55 [17]

Answer:

$\left[\begin{array}{cc}6&32\\39&15\\\end{array}\right]$ its the third option

Step-by-step explanation:

times each element by a half

8 0

3 years ago