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3 years ago

Which of these is closest in size to 1? 0.95 1.05 0.960 1.040 0.95

Mathematics

2 answers:

Artemon [7]3 years ago

7 0

Answer:

0.960

Step-by-step explanation:

andreyandreev [35.5K]3 years ago

6 0

Answer:

0.960

Step-by-step explanation:

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Write a recursive definition for the sequence 14, 10, 6, 2...

VladimirAG [237]

The general form of recursive definition for arithmetic sequence:
$a_{n}=a_{1}+d(n-1)$

So, what we need to do is to find $a_{1}$ and $d$ .
$a_{1}=14$
$d=a_{2}-a_{1}=10-14=-4$
Having found these two values we can now define our recursive sequence:
$a_{n}=14-4(n-1)$
$a_{n}=14-4n+4$
$a_{n}=18-4n$

7 0

3 years ago

Read 2 more answers

2/5 + 1/3, Whats The Answer?

meriva

Take the 2/5 and multiple it by 3 to get 6/15....then take the 1/3 and multiple it by 5 to get 5/15. Then add this together (6/15+5/15=....) to get 11/15

6 0

2 years ago

Read 2 more answers

Simplify 3 - (4x - 5) + 6.<br> A) 12x + 21 <br> B) 4x + 14 <br> C) 4x + 4 <br> D) -4x + 14

stepladder [879]

Answer:

D) -4x+14

Step-by-step explanation:

We are given with the expression

3 - (4x - 5) + 6

Here we are going to open the bracket at first step

as we have a negative sign in front of bracket the signs of the terms inside brackets will be reversed

3-4x+5+6

now we bring the like terms together

3+5+6 - 4x

Here 3, 5, 6 all are constant and hence like

14-4x

3 - (4x - 5) + 6= -4x+14

Hence our answer is -4x+14

6 0

3 years ago

(plz solve asap) find the value of x i dont know how to solve it so anyone can help

k0ka [10]

Answer:

I tried but sorry couldn't figure it out.

7 0

3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

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

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

3 years ago