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4 years ago

1/2+1/8 in simplest form

2 answers:

6 0

Find the Least Common Denominator (LCD) = 8

Make the denominators the same as the LCD 1*4 2/4 + 1/8

Simplify. Denominators are now the same = 4/8 + 1/8

Join the denominators = 4+1/8

Simplify = 5/8

Viktor [21]4 years ago

3 0

1/4 • 4/4 = 4/8 + 1/8 = 5/8 the dot means to multiply

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svetoff [14.1K]

Answer:

5+9+10+10+5+2+5+2

Step-by-step explanation:

48÷8=6

6 is median of given data

5 0

2 years ago

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A company buys computers and printers. Each computer costs $550 and each printer costs $390. If the company spends $8160 and buy

maxonik [38]

Answer:the company bought 12 computers and 4 printers.

Step-by-step explanation:

Let x represent the number of computers that the company bought.

Let y represent the number of printers that the company bought.

The company buys a total of 16 machines. It means that

x + y = 16

Each computer costs $550 and each printer costs $390. If the company spends $8160 for all the computers and printers that was bought, it means that

550x + 390y = 8160 - - - - - - - - - - 1

Substituting x = 16 - y into equation 1, it becomes

550(16 - y) + 390y = 8160

8800 - 550y + 390y = 8160

- 550y + 390y = 8160 - 8800

- 160y = - 640

y = - 640/ - 160

y = 4

Substituting y = 4 into x = 16 - y, it becomes

x = 16 - 4

x = 12

6 0

3 years ago

Anit [1.1K]

Answer:

-1/3

Step-by-step explanation:

(-2, 5) and (1, 4)

Slope:

m=(y2-y1)/(x2-x1)

m=(4-5)/(1+2)

m=(-1)/3

m= -1/3

5 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 X = number of catastrophes in the missions.

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

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X 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 n 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

Please Answer I’m desperate

igor_vitrenko [27]

Answer:

Step-by-step explanation:

Given

f = c / λ ( multiply both sides by λ )

fλ = c

4 0

3 years ago

Read 2 more answers