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mixer [17]
4 years ago
14

Im going to give 15 points for this question please answer it

Mathematics
1 answer:
Gnom [1K]4 years ago
8 0
I know, you have to divide 90 by what ever the length is, then you have to figure out where the points are. So it should be.... D!!!
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Answer is 345 make an equation from 100, 10, 3, 4, 5, 2, 1
Tpy6a [65]

Answer: 10 × 3 + 4 + 5 × 2 + 1 + 3 × 100 = 345

8 0
2 years ago
Please help, I’ve been stuck on this question for about a hour now.
Hoochie [10]
6x + 5x + (x + 16) + (3x - 1) = 360
11x + x + 16 + 3x - 1
15x + 15 = 360
15x = 345
x = 23


6(23) = 138
((23) + 16) = 39
(3(23) - 1) =68
5(23) = 115

138 + 39 + 68 + 115 = 360
7 0
3 years ago
Consider the following 8 numbers, where one labelled
AlladinOne [14]
<h3>Answers:     x = -17  and   x = 64</h3>

====================================================

Explanation

Consider three scenarios:

  • A) The value of x is the smallest of the set (aka the min)
  • B) The value of x is the largest of the set (aka the max)
  • C) The value of x is neither the min, nor the max. So 8 < x < 39.

These scenarios cover all the possible cases of what x could be. It's either the min, the max, or somewhere in between the min and max.

--------------------

We'll start with scenario A.

If x is the min, then that must mean 39 is the max as it's the largest of the set {18, 36, 16, 39, 27, 8, 34}

The range is 56, so,

range = max - min

56 = 39 - x

56+x= 39

x = 39-56

x = -17  which is one possible answer

--------------------

If instead we go with scenario B, then x is the max and 8 is the min

range = max - min

56 = x - 8

56+8 = x

64 = x

x = 64 is the other possible answer

--------------------

Lastly, let's consider scenario C. If x is not the min or the max, then it's somewhere between the min 8 and max 39. in short, 8 < x < 39.

Note that range = max - min = 39-8 = 31 which is not the range of 56 that we want. So there's no way scenario C can be possible here.

7 0
3 years ago
Help if you can plzz
melamori03 [73]

Answer:

you forgot the attachment

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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
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