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Delvig [45]
3 years ago
12

please helpppppppsoskskskjs

Mathematics
2 answers:
GREYUIT [131]3 years ago
7 0

Answer:

I am not sure but I am correct the simplified form of 12 :2 is 6:1. so 1:6/1

ss7ja [257]3 years ago
4 0

Answer:

1: 1/6 or you can say 1: 0.16

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HELP PLEASE ANSWER!!!!
Basile [38]

Answer: AAS

Step-by-step explanation:

Two sides are congruent, two angles are congruent, and vertical angles are congruent

5 0
3 years ago
In order to solve the following system of equations by subtraction, which of the following could you do before subtracting the e
Whitepunk [10]
I can't see the options, but you would need to multiply (4x - 2y = 7) by 3, and (3x - 3y = 15) by 4 so that you get 12x in both equations. Then when you subtract, you eliminate 12x. I hope this helps!
7 0
4 years ago
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Solve for y x+2y=6. what's the answer
Luda [366]

Solve for y

x+2y=6

Add (-x) to both sides:

x + 2y + (-x) = 6 + (-x)

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5 0
3 years ago
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Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime x (in weeks) has a gamma
elena-14-01-66 [18.8K]

Answer:

a) P(1 \leq X \leq 40)

In order to find this probability we can use excel with the following code:

=GAMMA.DIST(40;5,8,TRUE)-GAMMA.DIST(1,5,8,TRUE)

And we got:

P(1 \leq X \leq 40)=0.560

b) P(X \geq 40)=1-P(X

In order to find this probability we can use excel with the following code:

=1-GAMMA.DIST(40,5,8,TRUE)

And we got:

P(X \geq 40)=1-P(X

Step-by-step explanation:

Previous concepts

The Gamma distribution "is a continuous, positive-only, unimodal distribution that encodes the time required for \alpha events to occur in a Poisson process with mean arrival time of \beta"

Solution to the problem

Let X the random variable that represent the lifetime for transistors

For this case we have the mean and the variance given. And we have defined the mean and variance like this:

\mu = 40 = \alpha \beta  (1)

\sigma^2 =320= \alpha \beta^2  (2)

From this we can solve \alpha and [/tex]\beta[/tex]

From the condition (1) we can solve for \alpha and we got:

\alpha= \frac{40}{\beta}    (3)

And if we replace condition (3) into (2) we got:

320= \frac{40}{\beta} \beta^2 = 40 \beta

And solving for \beta = 8

And now we can use condition (3) to find \alpha

\alpha=\frac{40}{8}=5

So then we have the parameters for the Gamma distribution. On this case X \sim Gamma (\alpha= 5, \beta=8)

Part a

For this case we want this probability:

P(1 \leq X \leq 40)

In order to find this probability we can use excel with the following code:

=GAMMA.DIST(40;5,8,TRUE)-GAMMA.DIST(1,5,8,TRUE)

And we got:

P(1 \leq X \leq 40)=0.560

Part b

For this case we want this probability:

P(X \geq 40)=1-P(X

In order to find this probability we can use excel with the following code:

=1-GAMMA.DIST(40,5,8,TRUE)

And we got:

P(X \geq 40)=1-P(X

6 0
3 years ago
A circle has a radius of 48 millimeters. What is the central angle in radians, that intercepts an arc of length 36π millimeters?
Liono4ka [1.6K]
Arc length= radius x angle
36pie=48 x angle
divide both side by 48
you get 3/4pie
8 0
3 years ago
Read 2 more answers
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