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2 years ago

What is the value of x in the equation 3x -1/9 y = 18, when y = 27? 5 7 45 63

Mathematics

1 answer:

almond37 [142]2 years ago

4 0

Answer:

$x=7$

Step-by-step explanation:

we have

$3x-\frac{1}{9}y=18$

Substitute the value of y=27 in the equation and find the value of x

$3x-\frac{1}{9}(27)=18$

$3x-3=18$

$3x=18+3$

$3x=21$

$x=7$

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Step-by-step explanation:

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3 years ago

a beauty salon buys bottles of gel for $4.50 and marks up the price by 40% for what price do they sell each bottle?

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3 years ago

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Maggie only wore sandals tracker on two of the days that she trained so from her fitness trackers she verified that on the third

KIM [24]

We know form our problem that the third day she biked 20 miles, so we have the point (3,20). We also know that <span>on the eighth day she biked 35 miles, so our second point is (8,35).

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We can infer form our points that $x_{1}=3$ , $y_{1}=20$ , $x_{2}=8$ , and $y_{2}=35$ . so lets replace those values in our slope formula:
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$y-20=3(x-3)$
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$y=3x+11$

We can conclude that the equation of the line that best fit the set for Maggie’s data is $y=3x+11$ .

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3 years ago

What is the vertex of the graph of f(x) = x2 + 10x - 9? A) (-5, -34) B) (-5, -9) C) (5, -9) D) (5, 66)

sukhopar [10]

Answer:

A) (-5, -34)

Step-by-step explanation:

f(x) = x^2 + 10x - 9

We complete the square to get the equation in vertex form

Take the coefficient of the x term and divide by 2 then square it. We add it and then subtract it not to change the value of the equation

f(x) = x^2 + 10x +(10/2)^2 - (10/2)^2 - 9

f(x) = x^2 +10x +25 -25 -9

f(x) = (x^2 +10x +25) -34

The term in parentheses simplified to (x+10/2) ^2

= (x+5)^2 -34

= (x - -5)^2 -34

This is in the form (x-h)^2 +k

The vertex is (h,k) h=-5 and k=-34

(-5,-34)

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2 years ago

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Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
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Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

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3 years ago