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nika2105 [10]
3 years ago
6

Help quick A 1 B 2 C 3 D 4

Mathematics
2 answers:
tatiyna3 years ago
7 0
I believe 2 is the answer because it is an exact replica of abc
hope i helped

kati45 [8]3 years ago
6 0
The second one is the right answer. 
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Please help! what is y+7=-1/7(x+4) in slope intercept form?
Stells [14]
We know that
the equation of a line in <span>slope intercept form is--------------> y=mx+b
where
m-----------> is the slope
b-----------> is the y-intercept  point when x=0

</span><span>y+7=-1/7(x+4)---------> y+7=(-1/7)x-4/7
</span>y+7=(-1/7)x-4/7------> y=(-1/7)x-4/7-7------> y=(-1/7)x-(53/7)

the answer is
y=(-1/7)x-(53/7)-------------> this is the equation of a line in slope intercept form
m=(-1/7)=-0.14
b=(-53/7)=-7.57
y=-0.14x-7.57

see the attached figure

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3 years ago
Viyan age is 2 / 5 of the age of her eldest brother. If his brother's age is 35 years, then how old is viyan?
sashaice [31]
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6 0
3 years ago
Somebody help me with this please!!
Slav-nsk [51]

Notice the following pattern:

2 - 0 = 2

6 - 2  = 4

12 - 6 = 6

20 - 12 = 8

It's reasonable to assume that consecutive terms in the sequence differ by increasing multiples of 2, so that for the next number (call it x) we expect to see

x - 20 = 10  ==>  x = 30

and for the number after that (call it y) we would see

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3 years ago
Which of the following expressions has more than one term? (1 point)
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6 0
3 years ago
Read 2 more answers
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
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