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3 years ago

Alexander estimates the capacity of a small soup can to be about 1 fluid ounce. Find and correct his mistake

Mathematics

2 answers:

frez [133]3 years ago

8 0

Here we are given the small soup can is filled with 1 fluid ounce.

Also 1 fluid ounce can fill approximately 1/8 of a soup cup.

So its practically impossible to full a small soup cup with 1 fluid ounce.

Let us find out then how many fluid ounces can fill one soup cup.

Using unitary method:

If 1 fluid ounce can fill 1/8 of a cup.

Then to fill up the cup completely we need 8 fluid ounces.

yanalaym [24]3 years ago

5 0

The average size of soup would be around 8 fluid ounces.

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(1 point) A rectangular storage container with an open top is to have a volume of 12 cubic meters. The length of its base is twi

Vlada [557]

Answer:

The cost of the materials for the cheapest such container is approximately $249.8

Step-by-step explanation:

The volume the rectangular container is to have = 12 m³

The base length = 2 × The width of the base

The cost of the material for the base = 11 dollars/m²

The cost of the other sides = 9 dollars/m²

Let 'l', 'w', and 'h' represent the length and width of the base and the height of the rectangular storage container respectively, we have;

l = 2·w

l × w × h = 12 m³

∴ h = 12/(l × w) = 12/(2·w × w) = 6/w²

The cost of the open top rectangular container = The cost of the base + The cost of the 4 sides

The cost of the base = The area of the base × $11/m²

∴ The cost of the base = l × w × 11 =2·w × w × 11 = 22·w²

The cost of the 4 sides = The area of the four sides × $9/m²

∴ The cost of the 4 sides = 2 × (l × h + w × h) × 9

The cost of the 4 sides = 2 × (2·w × 6/w² + w × 6/w²) × 9 = 18 × (12/w + 6/w) = 324/w

∴ The cost of the open top rectangular container = 22·w² + 324/w

The coefficient of w² in the equation of the cost is positive, therefore, the cost of the materials for the cheapest such container, 'C', is given by value of 'w' at the minimum point of the equation, 22·w² + 324/w, which is given by the equating the derivative of the equation to zero as follows;

d(22·w² + 324/w)/dw = 0

∴ 44·w - 324/w² = 0

∴ 44·w = 324/w²

w³ = 324/44 = 81/11

w = ∛(81/11)

∴ C = 22·w² + 324/w = 22 × (∛(81/11))² + 324/(∛(81/11)) = 486/(∛(81/11)) ≈ 249.8

The cost of the materials for the cheapest such container, C ≈ $249.8

3 0

3 years ago

Instructions

S_A_V [24]

The theoretical probability of getting two tails on two coin tosses is 0.25.

<h3>How to calculate the probability?</h3>

The theoretical probability is the ratio of the number of favorable outcomes to the number of possible outcomes. Given a coin is tossed twice.

We have to find the theoretical probability of tossing two tails. The probability of getting tails on the toss of a coin is 1/2 0r 0.5.

Therefore, the probability of getting two tails on two coin tosses is 0.5 × 0.5 or 0.25.

The theoretical probability that a coin toss results in two heads showing is 0.25.

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

7 0

2 years ago

Anyone can y’all help me. What’s the mistake in this? And what’s N actually equal to?

PSYCHO15rus [73]

Answer:

n = -1

Step-by-step explanation:

you've accidently substracted 6n from 12n when you were supposed to be adding them. (-6n becomes +6n when brought to the otherside of the equal sign)

4 0

3 years ago

Someone plz answer this for me

pishuonlain [190]

Answer:

Step-by-step explanation:

D= $\sqrt{(1-(-5))^2+(4-(-4)^2}$

D= $\sqrt{6^2+8^2}$

D= $\sqrt{36+64}$

D= $\sqrt{100}$

D=10

3 0

3 years ago

An experiment consists of tossing a die and then flipping a coin once if the number on the die is even. If the number on the die

Mice21 [21]

Answer:

$\Omega = \{ 1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T,\\5HH , 5HT, 5TH, 5TT, 6H, 6T \}$

The probability is 3/12. The third option is correct.

Step-by-step explanation:

The sample space is

$\Omega = \{ 1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T,\\5HH , 5HT, 5TH, 5TT, 6H, 6T \}$

Note that this sample space is not equally probable.

The probability of getting a given number followed is the probability of getting an even number from the 6 numbers (3/6) multiplied by the probability of getting a head after getting that even number, that is 1/2, because is equally probable to get heads or tails from one single coin toss (note that we are assuming that the dice was even, thats why there is a single coin toss).

Therefore, the probability of getting an even number and a head is

P( D in {2,4,6} , H = 1) = P(D in {2,4,6}) * P(H=1 | D in {2,4,6}) = 3/6 * 1/2 = 3/12.

5 0

3 years ago