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san4es73 [151]
2 years ago
12

You just discovered that you have 100 feet of fencing and you have decided to make a rectangular garden. What is the largest are

a you can enclose using the materials you have? You must set up an equation and solve.
Mathematics
1 answer:
Rudik [331]2 years ago
8 0
The area is:
 A = x * y
 The perimeter is:
 P = 2x + 2y = 100
 We clear y:
 2y = 100-2x
 y = 50-x
 We write the area in terms of x:
 A (x) = x * (50-x)
 Rewriting:
 A (x) = 50x-x ^ 2
 Deriving:
 A '(x) = 50-2x
 We equal zero and clear x:
 50-2x = 0
 x = 50/2
 x = 25
 Then, the other dimension is given by:
 y = 50-x
 y = 50-25
 y = 25
 Therefore, the largest area is:
 A = (25) * (25)
 A = 625 feet ^ 2
 Answer:
 
the largest area you can enclose using the materials you have is:
 
A = 625 feet ^ 2
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The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar
Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

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The Foot Locker is having a 60% off sale on shorts. John paid $18 for a pair of shorts. What was the original price of the short
AURORKA [14]

Answer:

$30 is the price of the original pair of sneakers.

Step-by-step explanation:

You wanna cross multiply so you put :

18. 60

_ = _

x. 100

You multiply 18 the cost of the sneakers by 100 and then divide the product (1800) by 60 and you get $30.

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Juan invest $3700 in a simple interest account at a rate of 4% for 15 years
OleMash [197]

<em><u>Question:</u></em>

Juan Invest $3700 In A Simple Interest Account At A Rate Of 4% For 15 Years. How Much Money Will Be In The Account After 15 Years?

<em><u>Answer:</u></em>

There will be $ 5920 in account after 15 years

<em><u>Solution:</u></em>

<em><u>The simple interest is given by formula:</u></em>

S.I = \frac{p \times n \times r }{100}

Where,

p is the principal

n is number of years

r is rate of interest

From given,

p = 3700

r = 4 %

t = 15 years

Therefore,

S.I = \frac{3700 \times 4 \times 15 }{100}\\\\S.I = 37 \times 4 \times 15\\\\S.I = 2220

<em><u>How Much Money Will Be In The Account After 15 Years?</u></em>

Total money = principal + simple interest

Total money = 3700 + 2220

Total money = 5920

Thus there will be $ 5920 in account after 15 years

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The answer will be D. 7/10 because you have 10 numbers and you want to have a number greater than 3 so it would be 10-3=7 and 7 would go over 10 because there are 7 numbers greater than 3 but less than 10.

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