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dedylja [7]
3 years ago
9

Adding and subtracting polynomials

Mathematics
2 answers:
dedylja [7]3 years ago
4 0
This should be right

zzz [600]3 years ago
3 0

Answer:

The answer is 7x+4

Step-by-step explanation:

If you need any help just ask me.

I recommend  using math wa y calculator and I recommend any of the desmos calculators.

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5 fifths × 3 fifteinths =?
Pachacha [2.7K]

Answer:

1/5 or 0.2 i hope this helped you

8 0
3 years ago
Find the limit, if it exists. (if an answer does not exist, enter dne.) lim x → ∞ x4 x8 + 2
tatyana61 [14]

lim x → ∞ x^4 x^8 + 2

Combine exponents:

lim x → ∞ x^(4 +8) + 2

lim x → ∞ x^12 + 2

The limit at infinity of a polynomial, when the leading coefficient is positive is infinity.

7 0
3 years ago
For each sequence write an explicit formula 96,48,24,12,6
STALIN [3.7K]
a_1=96;\ a_2=48;\ a_3=24;\ a_4=12;\ a_5=6\\\\a_1=a_2:2\to96:2=48\\a_2=a_3:2\to48:2=24\\a_3=a_2:2\to24:2=12\\a_4=a_3:2\to12:2=6\\\\therefore\\\\a_n=96:2^{n-1}=\frac {96}{2^{n-1}}\\\\check:\\a_1=\frac{96}{2^{1-1}}=\frac{96}{1^0}=96\\a_2=\frac{96}{2^{2-1}}=\frac{96}{2^1}=\frac{96}{2}=48\\a_3=\frac{96}{2^{3-1}}=\frac{96}{2^2}=\frac{96}{4}=24\\\vdots
8 0
3 years ago
Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

P(X

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}

The cumulative distribution for this function is given by:

F(X) = 1- e^{-\lambda x}, x\ geq 0

We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

8 0
3 years ago
Harry ate 5/8 of a sandwich
11111nata11111 [884]
Well do you know a common factor between 8, 5, and 12, if so it is super simple to find your answer. :)
4 0
3 years ago
Read 2 more answers
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