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2 years ago

If y varies directly as x, and y is 18 when x is 5, which expression can be used to find the value of y when x is 11?

Mathematics

1 answer:

Leokris [45]2 years ago

5 0

Answer: The value of y is 39.6 when x = 11 and the required expression is y(11) = 39.6

Step-by-step explanation:

Since we have given that

y varies directly as x.

If y = 18, when x = 5.

So, we need to find the value of y when x is 11.

According to question, we get that

$\dfrac{y}{x}=\dfrac{18}{5}=\dfrac{y}{11}\\\\18\times 11=5y\\\\198=5y\\\\\dfrac{198}{5}=y\\\\y=39.6$

Hence, the value of y is 39.6 when x = 11 and the required expression is y(11) = 39.6

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Answer:

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2 years ago

Solve the system of equations using substitution

Andru [333]

Answer:

x = 136/11 , y = 68/11

Step-by-step explanation:

Solve the following system:

{6 x - y = 68

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Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for x:

{6 x - y = 68

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Hint: | Reverse the equality in 2 y = x in order to isolate x to the left hand side.

2 y = x is equivalent to x = 2 y:

{6 x - y = 68

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Hint: | Perform a substitution.

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{11 y = 68

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Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for y:

{11 y = 68

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Hint: | Solve for y.

Divide both sides by 11:

{y = 68/11

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Hint: | Perform a back substitution.

Substitute y = 68/11 into the second equation:

{y = 68/11

x = 136/11

Hint: | Sort results.

Collect results in alphabetical order:

Answer: {x = 136/11 , y = 68/11

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2 years ago

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3 years ago

A game is played with a spinner on a circle, like the minute hand on a clock. The circle is marked evenly from 0 to 100, so, for

zheka24 [161]

Answer:

The probability is 1/2

Step-by-step explanation:

The time a person is given corresponds to a uniform distribution with values between 0 and 100. The mean of this distribution is 0+100/2 = 50 and the variance is (100-0)²/12 = 833.3.

When we take 100 players we are taking 100 independent samples from this same random variable. The mean sample, lets call it X, has equal mean but the variance is equal to the variance divided by the length of the sample, hence it is 833.3/100 = 8.333.

As a consecuence of the Central Limit Theorem, the mean sample (taken from independant identically distributed random variables) has distribution Normal with parameters μ = 50, σ= 8.333. We take the standarization of X, calling it W, whose distribution is Normal Standard, in other words

$W = \frac{X - \mu}{\sigma} = \frac{X - 50}{8.333} \simeq N(0,1)$

The values of the cummulative distribution of the Standard Normal distribution, lets denote it $\phi$ , are tabulated and they can be found in the attached file, We want to know when X is above 50, we can solve that by using the standarization