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2 years ago

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Find the equation of the line that contains the points (3,5) and (7, 7). Write the equation in the form y = mx + b and identify

m and b.
m=
b=

1 answer:

Anastasy [175]2 years ago

8 0

Step-by-step explanation:

Using Point - Slope formula :

$(y - y_1) = \frac{(y_2 - y_1)}{(x_2 - x_1)} (x - x_1)$

$(y - 5) = \frac{2}{4} (x - 3)$

2y - 10 = x - 3

$y = \frac{1}{2} x + \frac{7}{2}$

m = 1 / 2

b = 7 / 2

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What is the value of the expression 2^0/4^2

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Anything to the zeroth power is equal to 1. (The numerator is equal to one because of this rule)

So.

$\frac{1}{4^2} = \frac{1}{16}$

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2 years ago

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Match each pair of equivalent fractions. 1 . 4/16 2 . 12/18 3 . 4/10 4 . 16/20

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Basically divide each fraction to get the simplified version of each answer. You divide each fraction by the highest multiple.
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2 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:

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

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2 years ago

The amount of money in a bank account increased by 15.8% over the last year. If the amount of money at the beginning of the year

Lostsunrise [7]

Answer:

The expression represents the amount of money in the bank account after the increase is: x=1.158n.

Step-by-step explanation:

With the information provided, you can say that the amount of money in the bank after the increase would be equal to multiplying the amount at the beginning of the year for the result of adding up 1 plus the percentage of the increase, which would be:

x=n*(1+0.158)

x=n*1.158

x=1.158n, where:

x is the amount of money in the bank account after the increase

n is the amount of money at the beginning of the year

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What's the question?

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