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densk [106]
3 years ago
8

Evaluate 13-0.75w+8x13−0.75w+8x13, minus, 0, point, 75, w, plus, 8, x when w=12w=12w, equals, 12 and x=\dfrac12x= 2 1 ​

Mathematics
1 answer:
Goryan [66]3 years ago
6 0

Answer:

the value of the given equation is 8

Step-by-step explanation:

The computation is shown below

Given that

13 - 0.75w + 8x

here w = 12

x = 1 ÷ 2

Now put these values to the above equation  

= 13 - 0.75 × (12) + 8 × (1 ÷ 2)

= 8

Hence, the value of the given equation is 8

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-2 + (3+x)/5 - 3 = -9 what is x
TiliK225 [7]

+2 on -2 and -9

+3 on -3 and 3

remove parenthesis

6/5 = 1.2

x= 1.2 / -.7

x=-1.4

7 0
2 years ago
Please help!! i have no idea!
Gemiola [76]

Answer:

θ = 60.34

Step-by-step explanation:

\frac{\left(12.8\cdot \:sin\left(90\right)\right)}{sin\left(52.3\right)} = 16.177

\frac{\left(16.177\cdot \:sin\left(90\right)\right)}{18.6} =  θ = .869

arcsin\left(.869\right) = 60.34

7 0
3 years ago
Read 2 more answers
Select three ratios that are equivalent to 8:20 Choose 3 answers: (A. 1:4
denpristay [2]

Answer:

8:20

divide by 2

B. 4:10

multiply by 3

C. 24: 60

divide by 4

D. 2:5

8 0
2 years ago
Solve for x: 8x-5= 6x + 1
Romashka-Z-Leto [24]

Step-by-step explanation:

8x-5= 6x + 1

8x - 6x = 1+5

2x = 6

x= 2/6 = 1/3

x= 0.333

6 0
3 years ago
Under average driving conditions, the life lengths of automobile tires of a certain brand are found to follow an exponential dis
wariber [46]

Answer:

a) P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

b) P(X>30000|X>15000)=P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

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}, x>0

And 0 for other case. Let X the random variable that represent "life lengths of automobile tires of a certain brand" and we know that the distribution is given by:

X \sim Exp(\lambda=\frac{1}{30000})

The cumulative distribution function is given by:

F(X) = 1- e^{-\frac{x}{\mu}}

Part a

We want to find this probability:

P(X>30000) and for this case we can use the cumulative distribution function to find it like this:

P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

Part b

For this case w want to find this probability

P(X>30000|X>15000)

We have an important property on the exponential distribution called "Memoryless" property and says this:

P(X>a+t| X>t)=P(X>a)  

On this case if we use this property we have this:P(X>30000|X>15000)=P(X>15000+15000|X>15000)=P(X>15000)

We can use the definition of the density function and find this probability:

P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

7 0
3 years ago
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