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Anni [7]
3 years ago
7

Help me solve this please

Mathematics
1 answer:
stealth61 [152]3 years ago
8 0

Answer:

the answer is 0.7692

Step-by-step explanation:

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What is the slope of a line that is perpendicular to the line whose equation is 8y−5x=11?
ElenaW [278]

9514 1404 393

Answer:

  -8/5

Step-by-step explanation:

When you solve for y, the slope of the line is the x-coefficient. For the given line, that is ...

  8y = 5x +11 . . . . . add 5x

  y = 5/8x +11/8 . . . . divide by 8

The given line has a slope of 5/8. The perpendicular line will have a slope that is the opposite of the reciprocal of this:

  -1/(5/8) = -8/5 . . . . slope of perpendicular line

4 0
3 years ago
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What formula is the same as B = atra?
horrorfan [7]

Answer:

atra=B

Step-by-step explanation:

that's the only one I can think of lol

3 0
3 years ago
Express the sum of 3x^2+5x-6 and -x^2+3×+9?
djverab [1.8K]
I hope this helps you



6 0
3 years ago
For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger
11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let <em>X</em> = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = p=\frac{1}{120}.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X </em>is:

P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...

In this case we need to compute the probability of 1 or more than 1 catastrophes in <em>n</em> missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419

6 0
3 years ago
What percent is represented by the shaded area?
kumpel [21]
95.5 percent is the answer I think
3 0
3 years ago
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