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3 years ago

2/5 of a kiddy pool was filled with water. After pouring out 5/7

Mathematics

1 answer:

Paladinen [302]3 years ago

8 0

Answer:

The question is asking what is the volume of water in the full pool. It is 70 liters.

Step-by-step explanation:

Let P represent the volume of a full pool.

"2/5 of a kiddy pool was filled with water" can be written as:

(2/5)P.

"After pouring out 5/7 of the amount of [I assume from the 2/5P] water in the pool" can be written as:

(2/7)(2/5)P, or (4/35)P [This represents the amount of water remaining in the pool after the above shenanigans.]

The we are told that when 62 liters are added back to the pool, it is full. That can be written as:

62 + (4/35)P = P [Add 62 liters to what was remaining. ((4/35)P), and we'll have P, a full pool, for a cool fool. [sorry]

62 = (31/35)P

P = 62(35/31)

P = 70 liters

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Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(O) = 2.8 y(t) =( Preview

Karolina [17]

Answer:

$y=\frac{4}{5}e^{t}-\frac{4}{5}e^{-\frac{2}{5}t}$

Step-by-step explanation:

The given equation 5y'' + 3y' - 2y =0 can be written as

$(5D^{2}+3D-2)y(t)=0$

Solving for complementary function we have Roots of $(5D^{2}+3D-2)$ as follows

$(5D^{2}+5D-2D-2)$

$5D(D+1)-2(D+1)=0\\\\(5D-2)(D+1)=0\\\\\therefore D=-1\\D=+2/5$

Thus the complementary function becomes

y= $y=c_{1}e^{m_{1}t}+c_{2}e^{m_{2}t}$

where

$m_{1},m_{2}$ are calculated roots

thus solution becomes

$y=c_{1}e^{-t}+c_{2}e^{\frac{2}{5}t}$

Now to solve for the coefficients we use the given boundary conditions

$y(0)=0\\\\\therefore c_{1}+c_{2}=0\\\\y'(0)=-c_{1}+\frac{2}{5}c_{2}=2.8\\\\\therefore c_{2}+\frac{2}{5}c_{2}=2.8\\\\c_{2}=2\\\\\therefore c_{1}=-2}$

hence the solution becomes

$y=-2e^-{t}+2e^{\frac{2}{5}t}$

8 0

3 years ago

Which number is a composite number 7 11 31 or 36 show your work.

Crank

36 because the rest are prime #'8

4 0

3 years ago

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After a large scale earthquake, it is predicted that 15% of all buildings have been structurally compromised.a) What is the prob

Westkost [7]

Answer:

a) 13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) 17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) 17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

d) 75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

e) The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

Step-by-step explanation:

For each building, there are only two possible outcomes after a earthquake. Either they have been damaged, or they have not. The probability of a building being damaged is independent from other buildings. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

15% of all buildings have been structurally compromised.

This means that $p = 0.15$

20 buildings

This means that $n = 20$

a) What is the probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) What is the probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised?

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0388 + 0.1368 = 0.1756$

17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) What is the probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised?

Either they find 4 or less, or they find more than 4. The sum of the probabilities of these events is 1. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293$

$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0388 + 0.1368 + 02293 + 0.2428 + 0.1821 = 0.8298$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.8298 = 0.1702$

17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

d) What is the probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised?

$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293$

$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X = 5) = C_{20,5}.(0.15)^{5}.(0.85)^{15} = 0.1028$

$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2293 + 0.2428 + 0.1821 + 0.1028 = 0.7570$

75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

e) What is the expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings?

The expected value of the binomial distribution is:

$E(X) = np$

$E(X) = 20*0.15 = 3$

The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

3 0

3 years ago

Order the measurements from largest to smallest. 4.9L , 4090ml , 1.4km , 4901m

Burka [1]

Answer:

4090 ml - 4.9L

1.4km - 4901m

Step-by-step explanation:

A liter is made up of 1000 ml so 4.9 l is equvilant to 4,900 ml.

A kilometre same as litre is made up of a 1000 m. So 1.4 km is 1400m

However, ml and l is a measurement of volume while m and km is a measurement of distance.

6 0

3 years ago

Which expression is equivalent to 6 x + 8?

miss Akunina [59]

Answer:

the 2nd one

Step-by-step explanation:

2*3x is 6x

2*4 is 8

equation is 6x+8

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3 years ago

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