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saveliy_v [14]
3 years ago
15

I need help now it’s an iready

Mathematics
1 answer:
schepotkina [342]3 years ago
3 0

answer:

21 + 3x (you would not put the x again since it is already given)

step-by-step explanation:

  • we have to distribute 3 to the expression in the parenthesis

3 (7 + x)

= 21 + 3x

  • this will be the expression once we distribute
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Y-intercept = (0, 2)<br> slope = -3/7 what is the equation?
Marianna [84]

Answer:

y = -(3/7)x + 2

Step-by-step explanation:

(see attached)

recall that the slope-intercept form of a linear equation is

y = mx + b

where m = slope = given as -(3/7)

and b = y-intercept = 2

substituting these values into the eqation:

y = mx + b

y = -(3/7)x + 2

8 0
3 years ago
Simplify the expression to a + bi form:<br> (9+7i)(6 + i)
Scorpion4ik [409]

(9 + 7i)(6 + i) \\  = 54 + 9i + 42i + 7 {i}^{2}  \\  = 54 + 51i  - 7 \\  = 47 + 51i

I hope I helped you^_^

6 0
3 years ago
Read 2 more answers
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
Item 8
SCORPION-xisa [38]

53 minutes

3:08 to 4:00 is 52 minutes.

to 4:01 would add an extra minute which would be 53.

6 0
3 years ago
Read 2 more answers
If your cell phone bill was $67.85 and there is a 7.5% late fee how much will your bill be?
dedylja [7]

67.85 \times .075 = 5.08875 \\ 67.85 + 5.08875 = 72.94
bill is $72.94
6 0
4 years ago
Read 2 more answers
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