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MA_775_DIABLO [31]
3 years ago
10

What is the opposite of running 150 feet east

Mathematics
2 answers:
ivolga24 [154]3 years ago
8 0

Answer:

running 150 feet west would be the opposite

SVETLANKA909090 [29]3 years ago
3 0

Answer:

running 150 feet west?

Step-by-step explanation:

is this the type of answer you wanted?

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Describe the types of transformation given by each parameter in g(x) = -4f(-x + 2) + 6
forsale [732]

Step-by-step explanation:

Starting with f(x)

f(x)

f(-x) ...... reflection in the y - axis

f(-x+2) ..... horizontal translation by 2 units in the negative x-direction

f(-x+2) + 6 ...... vertical translation by 6 units in the positive y-direction

4f(-x+2) + 6 ..... scaling (or dilation) by a factor of 4

-4f(-x+2) + 6 ..... reflection in the x-axis

7 0
4 years ago
The time T required to repair a machine is an exponentially distributed random variable with mean 10 hours.
Firdavs [7]

It can be expected about 36.79% of chance that repair time exceeds

The probability that a repair time exceeds  15 hours is 0.3679

What is the exponential distribution?

It explains about the time between events or the distance between two random events is termed the exponential distribution. Here, the occurrence of the events is continuous and also independent. Moreover, the average rate is constant.

The cumulative distribution function of T is obtained below:

From the information given, let the random variable T be the required time to repair a machine follows exponential distribution with parameter λ
with mean. 1/2 hours

That is,  E(x) =  1/2 hours.
The parameter of the random variable T is,
E(x) =  1/λ
λ = 1/E(x)
= 1/(1/2)
= 2

The probability density function of T is,
f(t) = \left \{ {{2e^{-2t} \ \ \ t > 0}  \  \atop {0} \ \ \ elsewhere} \ \right.
The cumulative density function of T is,
FT(t) = P(T <= t)

= 1 - e^{- \lambda t}

= 1 - e^{- 2t}
The CDF of T is,
P(T <= t)  = 1 - e^{- 2t}    0 <= T <= ∞
        = 0          otherwise
to obtain the probability that a repair time exceeds

1/2 hours.
(a) The probability that a repair time exceeds 1/2 hours.
From the given information, the CDF of T is,
P(T <= t)  = 1 - e^{- 2t}    0 <= T <= ∞
        = 0          otherwise

The required probability is,
P(T <= 1/2)  = 1 - P(T <= 1/2)
        = 1 - [  = 1 - e^{- 2(1/2)} ]
       = e^{-1}
= 0.3679

om total probability. It can be expected about 36.79% of chance that repair time exceeds

P(X => x)  = 1 - P(X < x)
to obtain the probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours.

(b), The probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours is obtained below:
From the given information, the CDF of T is,
P(T <= t)  = 1 - e^{- 2t}    0 <= T <= ∞
        = 0          otherwise
The required probability is,
P = P(T => 12.5∩T>12) / P(T>12)
= e^{- 25 + 24}

= e^{- 1}
= 0.3679
The probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours is obtained by dividing the
P = P(T => 12.5∩T>12) / P(T>12)
with
P(T>12).

It can be expected about 36.79% of chance that a repair takes at least 12.5 hours given that its duration exceeds 12 hours.

Hence, It can be expected about 36.79% of chance that repair time exceeds,

The probability that a repair time exceeds  15 hours is 0.3679

To learn more about the product of the fraction visit,
brainly.com/question/22692312
#SPJ4

7 0
1 year ago
The thickness of a book varies directly with
Bumek [7]

\qquad \qquad \textit{direct proportional variation}\\\\\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill }\\\\\textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill }\\\\[-0.35em]~\dotfill

\stackrel{\textit{"th" varies directly with "p"}}{th=kp}\qquad \textit{we also know that} \begin{cases} th=1.8\\ p=350 \end{cases} \\\\\\ 1.8=k(350) \implies \cfrac{9}{5}=350k\implies \cfrac{9}{1750}=k~\hfill \boxed{th=\cfrac{9}{1750}k} \\\\\\ \textit{when p = 84, what is "th"?}\qquad th=\cfrac{9}{1750}(84)\implies th=\cfrac{54}{135}\implies th=0.432

8 0
2 years ago
Which of the following are perpendicular to the line y=-1/3x+5 (three answers)
Mkey [24]
Y = -1/3x + 5....slope here is -1/3. A perpendicular line will have a negative reciprocal slope. All that means is " flip " the slope and change the sign. So ur perpendicular line will have a slope of 3.

y = 3x....has a slope of 3, so this is perpendicular.
3x - y = 2.....y = 3x- 2.....has a slope of 3 and is therefore perpendicular
3x - y = 6.....y = 3x - 6...has a slope of 3 and is perpendicular

6 0
3 years ago
What is the range of the data below <br> A.2 <br> B.5<br> C.12<br> D.13
Degger [83]

Range of the data = 13

Solution:

To find the range of the given data:

Let us first define what is range.

Range:

The range of the data set is the difference between the highest value and lowest value of the data set.

i. e. Range = Highest value – Lowest value

In the given number line,

Highest value indicated = 115

Lowest value indicated = 102

Range of the data = 115 – 102

                              = 13

Range of the data = 13

Hence the range of the given data is 13.

4 0
3 years ago
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