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Misha Larkins [42]
3 years ago
8

The ratio of the ages of three children is 2:4:5. The sum of their ages is 33. What is the age of each child?

Mathematics
2 answers:
sdas [7]3 years ago
4 0
We can represent this situation by the equation 2x+4x+5x=33.

Then 11x=33, and x = 3.  

Then the children's ages are in the ratio 2x:4x:5x, or 2(3):4(3):5(3), or

6:12:15.

Do these 3 ages add up to 33?  6 + 12 + 15 = 33?  YES!

The children's ages are 6, 12 and 15 years respectively.
Elanso [62]3 years ago
3 0

Step-by-step explanation:

let the given ages be 2x , 4x and 5x.

Here,

2x + 4x +5x =33

or,x = 33/ 11

or,x =3

Hence,

2x =2 (3)=6years

4x=4(3)=12years

5x=5(3)=15years

You might be interested in
6) Supplementary Exercise 5.51
tresset_1 [31]

Answer:

P(X \le 4) = 0.7373

P(x \le 15) = 0.0173

P(x > 20) = 0.4207

P(20\ge x \le 24)= 0.6129

P(x = 24) = 0.0236

P(x = 15) = 1.18\%

Step-by-step explanation:

Given

p = 80\% = 0.8

The question illustrates binomial distribution and will be solved using:

P(X = x) = ^nC_xp^x(1 - p)^{n-x}

Solving (a):

Given

n =5

Required

P(X\ge 4)

This is calculated using

P(X \le 4) = P(x = 4) +P(x=5)

This gives:

P(X \le 4) = ^5C_4 * (0.8)^4*(1 - 0.8)^{5-4} + ^5C_5*0.8^5*(1 - 0.8)^{5-5}

P(X \le 4) = 5 * (0.8)^4*(0.2)^1 + 1*0.8^5*(0.2)^0

P(X \le 4) = 0.4096 + 0.32768

P(X \le 4) = 0.73728

P(X \le 4) = 0.7373 --- approximated

Solving (b):

Given

n =25

i)

Required

P(X\le 15)

This is calculated as:

P(X\le 15) = 1 - P(x>15) --- Complement rule

P(x>15) = P(x=16) + P(x=17) + P(x =18) + P(x = 19) + P(x = 20) + P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)

P(x > 15) = {25}^C_{16} * p^{16}*(1-p)^{25-16} +{25}^C_{17} * p^{17}*(1-p)^{25-17} +{25}^C_{18} * p^{18}*(1-p)^{25-18} +{25}^C_{19} * p^{19}*(1-p)^{25-19} +{25}^C_{20} * p^{20}*(1-p)^{25-20} +{25}^C_{21} * p^{21}*(1-p)^{25-21} +{25}^C_{22} * p^{22}*(1-p)^{25-22} +{25}^C_{23} * p^{23}*(1-p)^{25-23} +{25}^C_{24} * p^{24}*(1-p)^{25-24} +{25}^C_{25} * p^{25}*(1-p)^{25-25}

P(x > 15) = 2042975 * 0.8^{16}*0.2^9 +1081575* 0.8^{17}*0.2^8 +480700 * 0.8^{18}*0.2^7 +177100 * 0.8^{19}*0.2^6 +53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0  

P(x > 15) = 0.98266813045

So:

P(X\le 15) = 1 - P(x>15)

P(x \le 15) = 1 - 0.98266813045

P(x \le 15) = 0.01733186955

P(x \le 15) = 0.0173

ii)

P(x>20)

This is calculated as:

P(x>20) = P(x = 21) + P(x = 22) + P(x = 23) + P(x = 24) + P(x = 25)

P(x > 20) = 12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1 +1 * 0.8^{25}*0.2^0

P(x > 20) = 0.42067430925

P(x > 20) = 0.4207

iii)

P(20\ge x \le 24)

This is calculated as:

P(20\ge x \le 24) = P(x = 20) + P(x = 21) + P(x = 22) + P(x =23) + P(x = 24)

P(20\ge x \le 24)= 53130 * 0.8^{20}*0.2^5 +12650 * 0.8^{21}*0.2^4 +2300 * 0.8^{22}*0.2^3 +300 * 0.8^{23}*0.2^2 +25* 0.8^{24}*0.2^1

P(20\ge x \le 24)= 0.61291151859

P(20\ge x \le 24)= 0.6129

iv)

P(x = 24)

This is calculated as:

P(x = 24) = 25* 0.8^{24}*0.2^1

P(x = 24) = 0.0236

Solving (c):

P(x = 15)

This is calculated as:

P(x = 15) = {25}^C_{15} * 0.8^{15} * 0.2^{10}

P(x = 15) = 3268760 * 0.8^{15} * 0.2^{10}

P(x = 15) = 0.01177694905

P(x = 15) = 0.0118

Express as percentage

P(x = 15) = 1.18\%

The calculated probability (1.18%) is way less than the advocate's claim.

Hence, we do not believe the claim.

5 0
3 years ago
If c (x) = 4x – 2 and d(x) = x2 + 5x, what is (cd) (x)
jolli1 [7]

c(x) = 4x - 2

d(x) = x^2 + 5x

4x - 2(x^2 + 5x)

<em><u>Using FOIL, distribute each term appropriately. </u></em>

4x^3 + 20x^2 - 2x^2 - 10x

<em><u>Combine like terms.</u></em>

4x^3 + 18x^2 - 10x is the simplified polynomials achieved when c(x) and d(x) are multiplied.

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Jay ate 2/3 of a pepperoni pizza. Darrell ate 2/4 of a mushroom pizza. If the pizza pans are the same size, who ate a greater fr
FinnZ [79.3K]

Answer:

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Step-by-step explanation:

Jay ate 2/3 of a pizza, while Darrell ate 2/4

2/3 is greater, like 6/9, but, 2/4 is only half, like 4.5/9

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