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Ilia_Sergeevich [38]
3 years ago
8

The original price for the sweater is $80. Today it is on sale for 30% off. What is the sale price for the sweater?

Mathematics
2 answers:
konstantin123 [22]3 years ago
8 0

Answer:

56

Step-by-step explanation:

30 percent of 80 is 24, subtract

omeli [17]3 years ago
3 0

Answer:

56

Step-by-step explanation:

8 x .7=56

you get .7 by subtracting the 30 percent or .3 from the original price

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In this question , it is given that

your pace on a treadmill is 40.0 meters per minute.

In means 40 meters are travelled in one minute .

And  1 meter equals to 3.28 feet .

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131.2 feets are travelled in 1 minute.

Therefore, 7220 feet are travelled in

= \frac{7220}{131.2} minutes = 55 minutes

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Answer:

0.1

Step-by-step explanation:

There are a total of 10 tiles, of which 1 is the letter G

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4 years ago
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Is 3/x+4/y=2 linear or non-linear
Eduardwww [97]

The given equation is non-linear.

Solution:

Given expression is \frac{3}{x} +\frac{4}{y} =2.

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Linear equation is a straight line which is in the form ax + by + c = 0.

Otherwise it is non-linear equation.

⇒ \frac{3}{x} +\frac{4}{y} =2

Do cross multiply for the fractions to make the denominators same.

⇒ \frac{3y}{yx} +\frac{4x}{yx} =2

⇒ \frac{3y+4x}{yx} =2

Do cross multiplication.

⇒ 3y+4x=2yx

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5 0
3 years ago
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What is the probability that e) A fair coin lands Heads 6 times in a row? f) A fair coin lands Heads 4 times out of 5 flips? g)
cricket20 [7]

Answer:

e) 1.56%

f) 15.62%

h) 0.879%

g) 11.72%

Step-by-step explanation:

What we will do is solve point by point.

e)  A fair coin lands Heads 6 times in a row?

We have the following:

Total number of possible outcomes = 2 ^ 6 = 64

Number of favorable outcomes = 1

Required probability = 1/64 = 1.56%

f) A fair coin lands Heads 4 times out of 5 flips

We have the following:

Total number of possible outcomes = 2 ^ 5 = 32

Number of favorable outcomes = 5C4

nCr = n! / (r! * (n-r)!)

5C4 = 5! / (4! * (5-4)!) = 5

Required probability = 5/32 = 15.62%

g) he bit string has exactly two 1s, given that the string begins with a 1 if you pick a bit string from the set of all bit strings of length ten?

We have the following:

Total number of possible outcomes = 2 ^ 10 = 1024

Number of ways in which a position excluding the start of the string can be chosen is 9C1

Total number of favorable outcomes = 9C1

9C1 = 9! / (1! * (9-1)!) = 9

Required probability = 9/1024  = 0.879%

h)The bit string has the sum of its digits equal to seven if you pick a bit string from the set of all bit strings of length ten?

We have the following:

Total number of possible outcomes = 2 ^ 10 = 1024

For the sum of the digits to be 7 there has to be 7 ones.

Number of ways in which 7 position can be chosen is 10C7.

Total number of favorable outcomes = 10C7

10C7 = 10! / (7! * (10-7)!) = 120

Required probability = 120/1024 = 11.72%

5 0
4 years ago
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