All categories

3 years ago

How to work out math problem %of 50 shirts is 35 shirts

2 answers:

7 0

50 shirts is equal to 100% of the shirts right? So, to find what percentage 35 shirts is, you can divide 35 by 50 to get the decimal value of the percentage. This gives you 0.7, which is the same as 70%.

35 shirts are 70% of 50 shirts.

lina2011 [118]3 years ago

6 0

"of' is multiplication.
"is" is equal.
So x% of 50 is 35 could be written as
x% $\times$ 50 = 35

The first step is to divide by 50 on both sides.
x% times 50 divided by 50 is x%.
35 divided by 50 is .7

So x% = .7.

The next step is to get rid of the percent sign to find the value of x (our answer). To do this, we multiply by 100 on both sides.
x% times 100 is x.
.7 times 100 is 70%.

So x = 70%.

The answer:
70% of 50 shirts is 35 shirts.

You might be interested in

If f(x)=3x^2 and g(x)=4x^3, what is the degree of (f*g)(x)?

Westkost [7]

3x^2 * 4x^3 = 12x^5

Answer is degree 5

3 0

3 years ago

Read 2 more answers

The Tire on franks bike moves 75 inches in one rotation. How many rotations will the tire have made after frank rides 50 feet

Sladkaya [172]

You have to convert 50 feet to inches, so you take 50 times 12 and get 600. Then, you take 600/75 to get the number of rotations. Your final answer shold be 8 rotations.

4 0

3 years ago

James has a balance of $1,230 on his credit card that has an APR of 24%. He currently pays the minimum monthly payment of

ArbitrLikvidat [17]

Answer:

the answer is B James can eliminate $25 from Food/Clothes and $27 from Entertainment.

hope it helps!

7 0

3 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago

Use the data set below to determine measurements of central tendency.

notka56 [123]

Mean = 40 median = 37.5 mode = 30

4 0

3 years ago