All categories

3 years ago

√3xy^3 • √8xy^4 = i need steps too.

Mathematics

2 answers:

Elodia [21]3 years ago

4 0

Answer:

2 x ² $y^{7}$ √ 6

Step-by-step explanation:

√3xy^3 • √8xy^4

Multiply x by x by adding the exponents.

$\sqrt{3}x^{2} y^{3} * (\sqrt{8} y^{4}$

Multiply y ^3 by y ^4 by adding the exponents.

liubo4ka [24]3 years ago

3 0

Step-by-step explanation:

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

2x²y exponit 7√6

You might be interested in

Write 51% as a fraction in simplest form

olga_2 [115]

51/100 is the answer

3 0

3 years ago

10 normal six sided dice are thrown.Find the probability of obtaining at least 8 failuresif a success is 5 or 6.

erastova [34]

Answer:

0.2992 = 29.92% probability of obtaining at least 8 failures.

Step-by-step explanation:

For each dice, there are only two possible outcomes. Either a failure is obtained, or a success is obtained. Trials are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A success is 5 or 6.

A dice has 6 sides, numbered 1 to 6. Since a success is 5 or 6, the other 4 numbers are failures, and the probability of failure is:

$p = \frac{4}{6} = 0.6667$

10 normal six sided dice are thrown.

This means that $n = 10$

Find the probability of obtaining at least 8 failures.

This is:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.6667)^{8}.(0.3333)^{2} = 0.1951$

$P(X = 9) = C_{10,9}.(0.6667)^{9}.(0.3333)^{1} = 0.0867$

$P(X = 10) = C_{10,10}.(0.6667)^{10}.(0.3333)^{0} = 0.0174$

Then

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1951 + 0.0867 + 0.0174 = 0.2992$

0.2992 = 29.92% probability of obtaining at least 8 failures.

8 0

3 years ago

Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. players win

Ivahew [28]

The rolls of the dice are independent, i.e. the outcome of the second die doesn't depend in any way on the outcome of the first die.

In cases like this, the probability of two events happening one after the other is the multiplication of the probabilities of the two events.

So, the probability of rolling two 6s is the multiplication of the probabilities of rolling a six with the first die, and another six with the second:

$P(\text{rolling two 6s}) = P(\text{rolling a 6}) \cdot P(\text{rolling a 6}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Similarly,

$P(\text{rolling two 3s}) = P(\text{rolling a 3}) \cdot P(\text{rolling a 3}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Actually, you can see that the probability of rolling any ordered couple is always 1/36, since the probability of rolling any number on both dice is 1/6:

$P(\text{rolling any ordered couple}) = P(\text{rolling the first number}) \cdot P(\text{rolling the second number}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$