Ask question

Ask question

All categories

3 years ago

15

1,000 shoes. 20% of them are sneakers. 40% of those sneakers are red.

2 answers:

Shkiper50 [21]3 years ago

8 0

Answer:80 red sneakers

Step-by-step explanation:

1000 shoes

20/100*1000=200 sneakers

40/100*200=80 red sneakers

fenix001 [56]3 years ago

4 0

20% of 1000 = 200

40% of 200 = 80

Therefore there are 80 red sneakers

Hope this helps :)

You might be interested in

Which statement about the quadratic equation below is true?

Vladimir79 [104]

Hmm, I'd say the answer is,

A) The equation has x = 4 as its only solution

3 0

3 years ago

My name is Ann [436]

I would say it’s the first one because if you round it it is going to be approximately 12 so that is my reason and also why it isn’t number two is because if you subtract 13.26 and 7.98 you get 5. Something so that’s why

4 0

3 years ago

Nitella [24]

Answer:

I hope this is the whole question

Step-by-step explanation:

Well I can only see one way and it is this...

$\sqrt{27a}$ = $\sqrt{3.3.3.a}$

= $3\sqrt{3a}$

I got rid of one perfect-square. I dont think you can do much more.

6 0

2 years ago

After all markdowns and discounts, Charlene's prom dress cost her $22 before tax. The dress was on a rack labeled "50% off lowes

uysha [10]

Answer: $88.

Step-by-step explanation:

given data:

The cost of the dress for the prom = $22

label of the rack the dresss was on = 50%.

lowest marked price for the dress = 75%.

Solution:

if the lowest marked price of the cloth was already 75%.

and the current cost of the clothe is $22

the original cost of the cloth would be

100% – 75% = 0.25%

= $22 * 100 / 0.25%

= $88.

The original cost of the dress was $88

7 0

4 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

3 years ago