All categories

3 years ago

a veterinarian purchases some medication for $19.50 and wants to make a $2 profit. what is the percent markup on this medication

Mathematics

1 answer:

MaRussiya [10]3 years ago

8 0

11%

I multiplied the price by percentages up until I got a little over 2.

You might be interested in

{(12*3)+6}-3^2 divided by (9-6)

Mrac [35]

33. 12*3=36 the u add 6 so it becomes 42. 3^2 is 9 so if u subtract 9 from 42 it becomes 33.

6 0

3 years ago

5c/f=1/r, solve for r, c and f.

Nikolay [14]

Answer:

c = 1, f = 5, and r = 1

Step-by-step explanation:

5(1)/(5) = 1/(1)

5/5 = 1/1

1 = 1

8 0

3 years ago

Read 2 more answers

What can you tell about the mean of each distribution?

Mars2501 [29]

Answer:

I'm sorry I don't know ‍♀️

4 0

3 years ago

Read 2 more answers

1/6 + 3/8 <br> A. 4/6<br> B. 4/8 <br> C. 4/14<br> D. 13/24

Naddika [18.5K]

Here is you're answer:

In order to get you're answer you need to find the common denominator then add.

$\frac{1}{6} + \frac{3}{8}$
Find the common denominator:
$CD = 48$
$= \frac{26}{48}$
Simplify:
$\frac{26}{48} \div 2 = \frac{13}{24}$
$= \frac{13}{24}$

Therefore you're answer is option D "13/24."

Hope this helps!

8 0

3 years ago

Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly sel

Vlad [161]

Answer:

The required probability is 0.94

Step-by-step explanation:

Consider the provided information.

There are 400 refrigerators, of which 40 have defective compressors.

Therefore N = 400 and X = 40

The probability of defective compressors is:

$\frac{40}{400}=0.10$

It is given that If X is the number among 15 randomly selected refrigerators that have defective compressors,

That means n=15

Apply the probability density function.

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

We need to find P(X ≤ 3)

$P(X\leq3) =P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\leq3) =\frac{15!}{15!}(0.1)^0(1-0.1)^{15}+\frac{15!}{14!}(0.1)^1(1-0.1)^{14}+\frac{15!}{13!2!}(0.1)^2(1-0.1)^{13}+\frac{15!}{12!3!}(0.1)^3(1-0.1)^{12}\\$

$P(X\leq3) =0.944444369992\approx 0.94$

Hence, the required probability is 0.94

4 0

3 years ago