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hoa [83]
3 years ago
14

Describe how to transform (^6 sqrt x^5)^7 into an expression with a rational exponent.

Mathematics
1 answer:
solniwko [45]3 years ago
5 0
\bf a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
\left(\sqrt[6]{x^5}\right)^7\implies \left( x^{\frac{5}{6}} \right)^7\implies x^{\frac{5}{6}\cdot 7} \implies x^{\frac{5\cdot 7}{6}}\implies x^{\frac{35}{6}}
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3 years ago
Simplify the expression. Write your answer as a power. 5^8 x 5^5/5^4 x 5^3
PSYCHO15rus [73]

Answer:

Step-by-step explanation:

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2 years ago
The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2885 and standard deviation
aliina [53]

Answer:

a.  X~N(2,885, 651)

b.  0.086291

c.  0.00058

d.  3213.10 calories

Step-by-step explanation:

a. -A normal distribution is expressed in the form X~N(mean, standard deviation).

-Let X a random variable denoting  the number of calories consumed.

-X is a is a normally distributed random variable with mean 2885 and standard deviation 651.

-This distribution is expressed as X~N(2,885, 651)

b. The probability that less than 2000 calories are consumed is calculated using the formula:

P(X

#substitute the given values in the formula to solve for P:

P(X

Hence, the probability of consuming less than 2000 calories is 0.08691

c. The proportion of customers consuming more than 5000 calories is calculated as:

P(X>x)=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(Z>\frac{5000-2885}{651})\\\\=P(z>3.2488)\\\\=1-0.99942\\\\=0.00058

Hence, the proportion of customers consuming over  5000 calories is 0.00058

d. The least amount of calories to get the award is calculated as:

1% is equivalent to a z value of 0.50399.

-We equate this to the formula to solve for the mean consumption:

0.01=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(z>\frac{\bar X-2885}{651})\\\\\1\%=0.50399 \\\\\frac{\bar X-2885}{651}=0.50399 \\\\\bar X=0.50399\times 651+2885\\\\=3213.09

Hence, the least amount of calories consumed to qualify for the award is 3213.10 calories.

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