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m_a_m_a [10]
3 years ago
6

I need help!! Can an awesome genius please help me out?​

Mathematics
2 answers:
SOVA2 [1]3 years ago
8 0

Answer:     The answer is D i hope this helps :)

Step-by-step explanation:

tensa zangetsu [6.8K]3 years ago
7 0

Answer:

D

Step-by-step explanation:

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PLEASE HELP! I’M SO CONFUSED!!
romanna [79]

Answer:

The answer is 7.

Step-by-step explanation:

4 divided  by 7 is 0.571428 these just keep repeating over and over again so the closest dividable number of 6 (6 is how many digits there are before it repeats) is 1998 so the 2nd digit in .571428 is 7 hope this helps.

8 0
3 years ago
What is 9×9 and 10×80
Damm [24]

9 x 9 = 81

10 x 80 = 800

3 0
3 years ago
Read 2 more answers
3 i
vova2212 [387]

x+1=-x+7

2x=6 /÷2

x=3 np

6 0
3 years ago
What is the equation of the quadratic graph with a focus of (5, −1) and a directrix of y = 1?
pav-90 [236]
(x-h)^2=4p(y-k)
vertex is (h,k)
p is the distance from focus to vertex and distance from vertex to directix (vertex is in middle of directix and focus)
if p is positive, the parabola opens up and the focus is above the directix
if p is negative, the parabola opens down and the focus is below the directix
 
we see the directix is over the focus (1>-1) so the parabola opens down and p is negative
distance from (5,-1) to y=1 is 2 units
2/2=1
p=-1 since p is negative
1 up from (5,-1) is (5,0)
veretx is (5,0)
(x-5)^2=4(-1)(y-0)
(x-5)^2=-4y is the equation
5 0
3 years ago
Read 2 more answers
The time between breakdowns of an alarm system is exponentially distributed with mean 10 days. What is the probability that ther
Nimfa-mama [501]

Answer:

P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

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

Solution to the problem

For this case the time between breakdowns representing our random variable T is exponentially distirbuted T \sim Exp (\mu = 10)

So on this case we can find the value of \lambda like this:

\lambda = \frac{1}{\mu} = \frac{1}{10}

So then our density function would be given by:

P(T)=\lambda e^{-\frac{t}{10}}

The exponential distribution is useful when we want to describe the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time between two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:

P(T>t)= e^{-\lambda t}

And on this case we are looking for this probability:

P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048

5 0
4 years ago
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