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2 years ago

HELPPPPPPPPPPP MEEE PLS

Mathematics

1 answer:

ivanzaharov [21]2 years ago

4 0

Answer:

See explanation below

Step-by-step explanation:

If the base of a radical exponent is the same then the following properties hole and x≠ 0

$\sqrt[n]{x}$ = $x^{\frac{1}{n}$ Rule 1

$\frac{x^m}{x^n} = x^{m-n}$ Rule 2

$(x^m)^n = x^{mn}$ Rule 3

The numerator in the expression is

$\sqrt[3]{x^2y^5} =(x^2)^\frac{1}{3} (y^5)^\frac{1}{3} = x^\frac{2}{3}} . y ^ \frac{5}{3}$ Using rules 1 and 3

The denominator is

$\sqrt[4]{x^3y^4} =(x^3)^\frac{1}{4} (y^4)^\frac{1}{4} = x^\frac{3}{4}} . y ^ 1}$ Using rules 1 and 3

Therefore the original expression in exponent form is

$\frac{x^\frac{2}{3} y ^ \frac{5}{3}}{x^\frac{3}{4} y ^ 1}}$

Using rule 2 we get the final expression as

$x^{\frac{2}{3}-\frac{3}{4}}$ $y^{\frac{5}{3}-1}$

2/3 - 3/4 = 8/12 - 9/12 = -1/2

5/3 - 1 = 5/3 - 3/3 = 2/3

So the final expression is

$x^{-\frac{1}{12}}y^{\frac{2}{3}}$

Proved

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3 years ago

If XX is a binomial random variable, compute the mean, the standard deviation, and the variance for each of the following cases:

Liula [17]

Answer:

(a) The mean, variance and standard deviation of X are 1.60, 0.96 and 0.98 respectively.

(b) The mean, variance and standard deviation of X are 3.20, 0.64 and 0.80 respectively.

(d) The mean, variance and standard deviation of X are 4.00, 0.80 and 0.89 respectively.

Step-by-step explanation:

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

The mean, variance and standard deviation of X are:

$\mu=np\\\sigma^{2}=np(1-p)\\\sigma=\sqrt{np(1-p)}$

(a)

For n = 4 and p = 0.40 compute the mean, variance and standard deviation of X as follows:

$\mu=np=4\times0.40=1.60\\\sigma^{2}=np(1-p)=4\times0.40\times(1-0.40)=0.96\\\sigma=\sqrt{np(1-p)}=\sqrt{0.96}=0.98$

Thus, the mean, variance and standard deviation of X are 1.60, 0.96 and 0.98 respectively.

(b)

For n = 4 and p = 0.80 compute the mean, variance and standard deviation of X as follows:

$\mu=np=4\times0.80=3.20\\\sigma^{2}=np(1-p)=4\times0.80\times(1-0.80)=0.64\\\sigma=\sqrt{np(1-p)}=\sqrt{0.64}=0.80$

Thus, the mean, variance and standard deviation of X are 3.20, 0.64 and 0.80 respectively.

(c)

For n = 3 and p = 0.50 compute the mean, variance and standard deviation of X as follows:

$\mu=np=3\times0.50=1.50\\\sigma^{2}=np(1-p)=3\times0.50\times(1-0.50)=0.75\\\sigma=\sqrt{np(1-p)}=\sqrt{0.75}=0.87$

Thus, the mean, variance and standard deviation of X are 1.50, 0.75 and 0.87 respectively.

(d)

For n = 5 and p = 0.80 compute the mean, variance and standard deviation of X as follows:

$\mu=np=5\times0.80=4.00\\\sigma^{2}=np(1-p)=5\times0.80\times(1-0.80)=0.80\\\sigma=\sqrt{np(1-p)}=\sqrt{0.80}=0.89$

Thus, the mean, variance and standard deviation of X are 4.00, 0.80 and 0.89 respectively.

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3 years ago

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Anna007 [38]

Answer:

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Step-by-step explanation:

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3 years ago

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4 years ago

4+ y = 10 What is the answer and how do I solve it

julia-pushkina [17]

Answer:

y=6

Step-by-step explanation:

4+y=10

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3 years ago