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Svetach [21]
4 years ago
8

First is brainliest 0~0

Mathematics
1 answer:
egoroff_w [7]4 years ago
6 0

Answer:

65.60

Step-by-step explanation:

First find the markdown

markdown = original price times the percent markdown

                  = 80* 18%

                  = 80 * .18

                     14.4

Take the markdown from the original price to find the retail price

80-14.4 =65.60

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Given a feedlot of about 1200 cows, if an individual cow produces about 50 liters of manure a day, how many liters of manure wou
Amiraneli [1.4K]
Hello!

First you have to find how many liters of manure is made a day

1 cow makes 50 liters a day

We can do 50 * 1200 to find how many liters the cows make a day

50 * 1200 = 60000

Then you multiply this my how many days are in a month

60000 * 30 = 1800000

The answer is 1,800,000 liters of manure a month

Hope this helps!
3 0
4 years ago
8th grade math, please help.
Anastasy [175]

Answer:

V = 15/4π = 3.75π

Step-by-step explanation:

V=πr^2h

V=π(1/2^2)15

V=π(1/4)15

V=15/4π

5 0
3 years ago
What is 1.1 as a percent
katrin2010 [14]
1.1 = 110% hope this helps!
7 0
4 years ago
Read 2 more answers
Which of the following is the function F(x) if F^-1(x) = x+4/11?
Marianna [84]

Answer:

B. F(x) =X-4/11

Step-by-step explanation:

We are given the inverse function;

f^(-1)(x) = x + 4/11

Thus means, x was replaced with y and vise versa. Thus;

x = y + 4/11

Lets make y the subject;

y = x - 4/11

Thus,in inverse of a function we replace y with f(x) for the original function or f^(1)(x) for the inverse.

Thus, putting f(x) for y gives;

f(x) = x - 4/11

8 0
3 years ago
Assume that the population proportion is 0.46. Compute the standard error of the proportion, σp, for sample sizes of 500,000; 1,
sladkih [1.3K]

Answer:

a) 0.00070

b) 0.00050

c) 0.00022

d) 0.00016

e) 0.00005

Step-by-step explanation:

Standard error for proportion formula

S.E = √P(1 - P)/n

Where P = proportion

n = number of samples

Assume that the population proportion is 0.46. Compute the standard error of the proportion, σp, for sample sizes of a) 500,000

S.E = √P(1 - P)/n

= √0.46 × 0.54/500000

= √ 4.968 ×10^-7

= 0.0007048404

≈ 0.00070

b) 1,000,000

√P(1 - P)/n

= √0.46 × 0.54/1000000

= 0.0004983974

≈ 0.00050

c) 5,000,000

√P(1 - P)/n

= √0.46 × 0.54/5000000

= √ 4.968 ×10^-8

= 0.0002228901

≈ 0.00022

d) 10,000,000

√P(1 - P)/n

= √0.46 × 0.54/10000000

= √2.484 ×10^-8

= 0.0001576071

≈ 0.00016

e) 100,000,000

√P(1 - P)/n

= √0.46 × 0.54/100000000

= √2.484 × 10^-9

= 0.0000498397

= 0.00005

4 0
3 years ago
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