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3 years ago

a local middle school has 99 computers and 333 students . What is the number of students per computer ?

1 answer:

7 0

It would be a decimal which is 3.363636363636364 and if you convert it into a fraction it is 9090909091/25000000000 so in real terms it is 3 students per computer.

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Solve for x .problem down below

AleksAgata [21]

A: no solution!

first, simplify each side of the equation.

3x + 5 - 10x simplifies to -7x + 5.
8 - 7x - 12 simplifies to -7x - 4.

then, add +7x on both sides of the equation to get the variable alone. if you add 7x to each side, you get left with 0.

so, that leaves 5 = -4 which is not true. so, that means there is no solution.

5 0

3 years ago

For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability of obtaining a score less than or equal to 12.
∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

7 0

4 years ago

Factor COMPLETELY the expression 2x^3 - 250

AURORKA [14]

Answer:

2 (x - 5) (x^2 + 5 x + 25)

Step-by-step explanation:

Factor the following:

2 x^3 - 250

Factor 2 out of 2 x^3 - 250:

2 (x^3 - 125)

x^3 - 125 = x^3 - 5^3:

2 (x^3 - 5^3)

Factor the difference of two cubes. x^3 - 5^3 = (x - 5) (x^2 + x 5 + 5^2):

2 (x - 5) (x^2 + 5 x + 5^2)

5^2 = 25:

Answer: 2 (x - 5) (x^2 + 5 x + 25)

5 0

4 years ago

Compute the missing data in the table for the following exponential function . Leave your answer in decimal form.

Snezhnost [94]

Answer: Option C: 0.03125

Step-by-step explanation:

we have an exponential equation, so we have:

f(x) = A*r^x

we have that:

f(1) = A*r^1 = A*r = 0.5

f(2) = A*r^2 = 0.25

f(3) = A*r^3 = 0.125

f(4) = A*r^4 = 0.0625

f(5) = A*r^5 = ?

we want to find the value f(5)

first, let's take the quotient:

f(2)/f(1) = (A*r^2)/(A*r) = r = 0.25/0.5 = 0.5.

then we have:

f(x) = A*0.5^x

and f(1) = A*0.5 = 0.5

then A = 0.5/0.5 = 1.

now we know that our function is f(x) = 0.5^x.

then, if we want to find f(5) we have:

f(5) = 0.5^5 = 0.03125

Then the correct option is option c.

4 0

3 years ago

Identify the terms and like terms in the expression.

Nesterboy [21]

9x and 5x are "like terms."

9x - 5x produces the same result as does -5x + 9x.

4 0

4 years ago