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4 years ago

Solve the equation for the specified variable S=gr-5gk*3, for g

Mathematics

1 answer:

Aleks [24]4 years ago

8 0

Answer:

$g = \sqrt{\frac{S}{3kr^{-5}}}$

Step-by-step explanation:

Given

$S=gr^{-5}gk*3$

Required

Solve for g

$S=gr^{-5}gk*3$

Split into different entities

$S=g * r^{-5}*g*k*3$

Reorder in terms of like terms

$S=g*g * r^{-5}*k*3$

$S=g^2 * r^{-5}*k*3$

$S=g^2 * 3kr^{-5}$

Solve for $g^2$

$g^2 = \frac{S}{3kr^{-5}}$

Take square root of both sides

$g = \sqrt{\frac{S}{3kr^{-5}}}$

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-2/3-5/6= write in simplest form

sleet_krkn [62]

-2/3 - 5/6

We need common denominators so, since 3 can go into 6, we only need to multiply the first fraction by 2.

= -2 x 2 / 3x2 - 5/6
= -4 / 6 - 5/6

We only subtract the numbers that are in the numerators,

= -4-5 / 6
= -9/6

Both nine and six are divisible by 3, so to put into lowest terms...

=-9÷3 / 6÷3
= -3/2 <--- Final Answer

3 0

3 years ago

5. given ABC, find the value of x

Fudgin [204]

There's no question g

6 0

3 years ago

If three pears and four oranges cost $4.85 and three pears and ten oranges cost $8.75 what is the cost of a pear and an orange

marishachu [46]

Answer:

1 pear = $0.75; 1 orange = $0.65

Step-by-step explanation:

(1) 3P + 4O = 4.85

(2) 3P + 10O = 8.75 Eqn (2) - Eqn (1)

3P + 10O – 3P – 4O = 8.75 – 4.85 Combine like terms

6O = 3.90 Divide each side by 6

O = $0.65 Substitute into Eqn (1)

3P + 4×0.65 = 4.85

3P + 2.60 = 4.85 Subtract 2.60 from each side

3P = 2.25 Divide each side by 3

P = $0.75

Oranges cost $0.65 each and pears are $0.75 each

4 0

3 years ago

Student records suggest that the population of students spends an average of 6.30 hours per week playing organized sports. The p

Ymorist [56]

Answer:

a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Step-by-step explanation:

To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .