CAN SOMEONE HELP ME PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

Mathematics

1 answer:

Anettt [7]3 years ago

7 0

The LCD is the product of (x -2)(x + 1)
To clear the fractions, multiply each term by the part of the LCD that the denominator is missing.
x(x + 1) + (x - 2)(x - 1) = -1(x - 2)(x + 1)
Distribute and foil, then combine like terms.
x^2 + x + x^2 - 3x + 2 = -x^2 + x + 2
2x^2 - 2x + 2 = -x^2 + x + 2
add x^2 to both sides
3x^2 -2x + 2 = x + 2
subtract from both sides
3x^2 - 3x + 2 = 2
subtract 2 from both sides.
3x^2 - 3x = 0
Factor out the GCF
3x(x - 1) = 0
Set each factor equal to zero and solve.
3x = 0 x - 1 = 0
x = 0 x = 1

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natta225 [31]

Answer:

$\displaystyle V = 2144.66 \ ft^3$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Volume of a Sphere Formula: $\displaystyle V = \frac{4}{3} \pi r^3$

r is radius

Step-by-step explanation:

Step 1: Define

Identify variables

r = 8 ft

Step 2: Find Volume

Substitute in variables [Volume of a Sphere Formula]: $\displaystyle V = \frac{4}{3} \pi (8 \ ft)^3$
Evaluate exponents: $\displaystyle V = \frac{4}{3} \pi (512\ ft^3)$
Multiply: $\displaystyle V = \frac{2048}{3} \pi \ ft^3$
Estimate: $\displaystyle V = 2144.66 \ ft^3$

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3 years ago

Help binomial distributions

Ymorist [56]

We have the formula to compute the probability of having exactly k successed over n trials, given a probability p of success (and implicitly a probability 1-p of failure), which is

$P(\text{k successed on n trials}) = \binom{n}{k}p^k(1-p)^{n-k}$

Now, the probability of at least 3 successes is the union of the following event: exactly three successes,exactly four successes and exactly five successes.

We can compute their probability and sum them: $\binom{5}{3}\left(\frac{3}{7}\right)^3\left(\frac{4}{7}\right)^2 + \binom{5}{4}\left(\frac{3}{7}\right)^4\left(\frac{4}{7}\right)^1 + \binom{5}{5}\left(\frac{3}{7}\right)^5 \approx 0.36788$