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Alexxx [7]
2 years ago
8

Simplify the radical expression.

Mathematics
2 answers:
Liula [17]2 years ago
6 0

Answer:

xy^3\sqrt[4]{y^3}

Step-by-step explanation:

\sqrt[4]{x^4y^{15}} =\sqrt[4]{x^4}*\sqrt[4]{y^{15}}\\\\=x\sqrt[4]{y^{15}}\\\\=x(\sqrt[4]{y^{12}} *\sqrt[4]{y^3})\\\\=xy^3\sqrt[4]{y^3}\\\\\\\sqrt[4]{y^{12}} =y^{12/4}=y^3 \\\\\sqrt[4]{x^4} =x^{4/4}=x

ankoles [38]2 years ago
5 0

Answer:

xy^3 \sqrt[4]{y^3}

Step-by-step explanation:

Recall the exponent property a^b+a^c=a^{(b+c)}.

We can use this property to break the problem down:

\sqrt[4]{x^4y^{15}}=\sqrt[4]{x\cdot x\cdot x\cdot x\cdot y^3\cdot y^3\cdot y^3\cdot y^3\cdot y^3}=\boxed{xy^3 \sqrt[4]{y^3}}

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Read 2 more answers
A computer maker receives parts from three suppliers, S1, S2, and S3. Fifty percent come from S1, twenty percent from S2, and th
rewona [7]

Answer:

a) 4.9 % of all parts is defective or 0.049 of the total parts.

b)  0.5102 is the probability that the defective part was supplied by S1

Step-by-step explanation:

N is the total number of parts from supplier S1, S2 and S3.

N1 = 0.5*N is the total number of part supplied by S1

N2 = 0.2*N is the total number of part supplied by S2  

N3 = 0.3*N is the total number of part supplied by S3

a) if Nd1 is the number of defective parts from supplier S1, Nd2 is the number of defective parts from supplier S2 and Nd3 is the number of defective parts from supplier S3, the the total defective parts Nd is:

Nd = Nd1 + Nd2 + Nd3, where

Nd1 = 0.05*N1 = 0.05*0.5*N = 0.025*N,

Nd2 = 0.03*N2 = 0.03*0.2*N = 0.006*N,

Nd3 = 0.06*N3 = 0.06*0.3*N = 0.018*N,

Then Nd = Nd1 + Nd2 + Nd3 =  0.049*N, so Nd/N = 0.049

b) P(S1 \vert d) = \frac{P(S1,d)}{P(d)} = \frac{P(d \vert S1)}{P(d)} = \frac{0.05*0.5}{0.049} \approx 0.5102

for the last expression I used the Bayes tehorem.

P(S1 \vert d) is the probability that occur S1 given that d (defective) is true. This a conditional probability.

see at https://en.wikipedia.org/wiki/Bayes%27_theorem

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