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3 years ago

The shoes are on sale for 39.99. If they were marked down 20%, what was the original price?

Mathematics

1 answer:

Damm [24]3 years ago

8 0

Answer:

32 US dollars

Step-by-step explanation:

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Use the given graph to determine the limit, if it exists.

Galina-37 [17]

<span>"A coordinate graph is shown with a downward sloped line crossing the y axis at the origin that ends at the open point 3, -1" → this is the left side of the graph, so the limit is -1.</span>

6 0

3 years ago

Help...please?? Thanks guys

vfiekz [6]

Answer:

x=20

Step-by-step explanation:

The Pythagorean theorem is

a^2 +b^2 = c^2

where a and b are the legs and c is the hypotenuse.

The legs are x and 15 and the hypotenuse is 25

x^2 + 15^2 = 25^2

x^2 +225 = 625

Subtract 225 from each side

x^2 +225-225 = 625-225

x^2 = 400

Take the square root of each side

sqrt(x^2) = sqrt(400)

x = 20

4 0

3 years ago

Read 2 more answers

A Nevada roulette wheel had 38 pockets, labeled as 0 00 1 2 3 ... 35 36. One bet is single number. If you bet $1 on a number (an

slavikrds [6]

Answer:

-0.0526

Step-by-step explanation:

Let X be the random variable denoting the net gain(in dollars) for a single trial(one bet).

Assuming that each number in the wheel is equally likely, probability of the outcome being a victory is $\frac{1}{38}$ and probability of failure is $\frac{37}{38}$ . For a win, X takes value 35 and for a loss X takes value -1. So the model is,

$P(X=35) = \frac{1}{38}$

$P(X=-1) = \frac{37}{38}$

$P(X=i) = 0 \;if\; i \neq 38, i\neq -1$

The mean for one bet is $E(X) = \sum xP(X=x) = 35\times\frac{1}{38} - 1\times\frac{37}{38} = -0.0526$

3 0

3 years ago

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

8 0

3 years ago

Two brothers went shopping at a back to school sale where all shirts and shorts were the same price. The younger brother spent $

Lana71 [14]

145÷14=10.357142857142857143

5 0

3 years ago