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2 years ago

In the equation (m^5/6) (m^1/6)^7 = m^x , what is the value of x that makes the equation true?

Mathematics

1 answer:

vlabodo [156]2 years ago

6 0

${m}^{ \frac{5}{6} } \times (m {}^{ \frac{1}{6} } ) {}^{7 } = m {}^{x}$

$m {}^{ \frac{5}{6} } \times m {}^{ \frac{7}{6} } = m {}^{x}$

${m}^{ \frac{5 \times 7}{6 \times 6} } = {m}^{x}$

$\frac{35}{36} = x$

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Using approciate properties find -4/7*3/5+4/7*-1/5+2/5*-4/5

slega [8]

Answer: 0.548

Step-by-step explanation:

First you need to remember the formula for PEMDAS(Parenthesis,exponents,multiplication,divison,addition,subtractio)

you can do either multiplication or divison whichever comes first, same with addition+subtraction.

the underlined means to simplify

1.-0.57x3/5+4/7x-1/5+2/5x-4/5

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3.-0.342+4/7x-1/5+2/5x-4/5

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5.-0.342+0.57x-0.2+2/5x-4/5

6.-0.342+0.114+2/5x-4/5

7.-0.342+0.114+0.4x-4/5

8.-0.342+0.114+-1.6/5

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3 years ago

A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses

Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = average age of the random sample of horses with colic = 12 yrs

$\mu$ = average age of all horses seen at the veterinary clinic = 10 yrs

$\sigma$ = standard deviation of all horses coming to the veterinary clinic = 8 yrs

n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P( $\bar X$ $\geq$ 12)

P( $\bar X$ $\geq$ 12) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{12-10}{\frac{8}{\sqrt{60} } }$ ) = P(Z $\geq$ 1.94) = 1 - P(Z < 1.94)

= 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

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3 years ago

A chemical company makes two brands of antifreeze. The first brand is 20% pure antifreeze, and the second brand is 70% pure anti

andre [41]

Answer:

First brand of antifreeze: 21 gallons

Second brand of antifreeze: 9 gallons

Step-by-step explanation:

Let's call A the amount of first brand of antifreeze. 20% pure antifreeze

Let's call B the amount of second brand of antifreeze. 70% pure antifreeze

The resulting mixture should have 35% pure antifreeze, and 30 gallons.

Then we know that the total amount of mixture will be:

$A + B = 30$

Then the total amount of pure antifreeze in the mixture will be:

$0.2A + 0.7B = 0.35 * 30$

$0.2A + 0.7B = 10.5$

Then we have two equations and two unknowns so we solve the system of equations. Multiply the first equation by -0.7 and add it to the second equation:

$-0.7A -0.7B = -0.7*30$