If you have 500 dollars and spend 20 every week then what would be the slope explain

A suit is on sale for $154. If the original price of the suit was $440, what percent was discounted for the sale?

9966 [12]

The dollar amount of the discount is 440 - 154 = $286.
Therefore the percentage discount for the sale is given by:
$\frac{286}{440}\times100=65\ percent$

3 0

3 years ago

My cousin goes to a strange schools. It's called Geometric Middle School. My cousin's math project is to determine how many tile

Anon25 [30]

Area of trapezoid = (1/2) (b1 + b2) h
= (1/2) x (12 + 24) x 6 sqrt 3
= 108 sqrt 3

area of Hexagon = 2 x Area of trapezoid
= 2 x 108 sqrt 3
= 216 sqrt 3

4 0

2 years ago

A student's grade for the four quarters are 95, 83, 91, and 85. What does the student need to earn on the final exam to have an

gulaghasi [49]

Answer:

The least score the student can get is 71 to get an overall B average at the end of the year.

Step-by-step explanation:

Let the score she needs to get be = x

We now have to take the mean of all the five subjects (including x) and equate it to the least score she can have to get a B grade.

Let us take the least score the student needs to get a B grade to be 85

Mean of all 5 scores =

$\frac{95+83+91+85+x}{5}=85$

$\frac{354+x}{5} =85$

$354+x = 425$

$x = 71$

The least score the student can get is 71 to get an overall B average at the end of the year.

3 0

2 years ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

Answer-

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

Solution-

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

6 0

2 years ago

One hundred items are simultaneously put on a life test. Suppose the lifetimes

romanna [79]

Answer:

a) $\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}$

b) $\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}$

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( $i-1$ )st and the $ith$ failures. Then, the $T_{i}$ are independent with $\mathrm{T}_{\mathrm{i}}$ being exponential with rate $\frac{(101-i)}{200} .$