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2 years ago

if you had a coupin for 12% off a meal & if you were charged 11$ what was the original price of the meal?

1 answer:

3 0

$12.50

Set up a ratio.
If it was 12% off, you paid 88%.
So 88/100=11/x
Cross multiply and solve for x.
So x=1100/88=12.50

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Help? With this question please

Pie

A. original stuff is for 6 ppl anf u need for 2. and since 2 is 3x less yhan six. divide the ingrefiants by 3.

b. for 15 ppl, 6 is default. dou le all ingtediants is for 12. thrn add 50% more. aka mulyiply all ingrediants by 2.5

6 0

2 years ago

Read 2 more answers

Write an equation in slope-intercept form for the line that is parallel to the given line and that passes through the given poin

Mamont248 [21]

Y=3x+4 is the equation you are looking for!

8 0

2 years ago

If x represents a number, does 2/5 times x always represent 40% of that number?

marusya05 [52]

Answer:

Yes, it always represent 40% of that number.

Step-by-step explanation:

5 0

2 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

2 years ago

The data shows the speed, in miles per hour, of motorists on a stretch of road.

melomori [17]

Answer:

i. 18 miles per hour is an outlier

ii. the outlier decreases the mean speed

Step-by-step explanation:

An outlier in a given data is one of the values that is far greater or lesser compared to others. It affect the mean and standard deviation of a given data significantly.

From the given data, 18 is far too small compared to other values. This is certainly an outlier. This would affect the mean speed by decreasing the value.

An interquartile range is a measure of differences among data by dividing a set of given data into quartile. Increasing the value of the outlier would increase the interquartile range.

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7 0

2 years ago