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Nana76 [90]
3 years ago
7

A certain number of sixes and nines is added to give a sum of 126; if the number of sixes and nines is interchanged, the new sum

is 114. How many of each were there originally?
Mathematics
1 answer:
galina1969 [7]3 years ago
7 0

Answer:

Original number of sixes = 6

Original number of nines = 10

Step-by-step explanation:

We are told in the question that:

A certain number of sixes and nines is added to give a sum of 126

Let us represent originally

the number of sixes = a

the number of nines = b

Hence:

6 × a + 9 × b = 126

6a + 9b = 126.....Equation 1

If the number of sixes and nines is interchanged, the new sum is 114.

For this second part, because it is interchanged,

Let us represent

the number of sixes = b

the number of nines = a

6 × b + 9 × a = 114

6b + 9a = 114.......Equation 2

9a + 6b = 114 .......Equation 2

6a + 9b = 126.....Equation 1

9a + 6b = 114 .......Equation 2

We solve using Elimination method

Multiply Equation 1 by the coefficient of a in Equation 2

Multiply Equation 2 by the coefficient of a in Equation 1

6a + 9b = 126.....Equation 1 × 9

9a + 6b = 114 .......Equation 2 × 6

54a + 81b = 1134 ........ Equation 3

54a + 36b = 684.........Equation 4

Subtract Equation 4 from Equation 3

= 45b = 450

divide both sides by b

45b/45 = 450/45

b = 10

Therefore, since the original the number of nines = b,

Original number of nines = 10

Also, to find the original number of sixes = a

We substitute 10 for b in Equation 1

6a + 9b = 126.....Equation 1

6a + 9 × 10 = 126

6a + 90 = 126

6a = 126 - 90

6a = 36

a = 36/6

a = 6

Therefore, the original number of sixes is 6

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Answer:

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Step-by-step explanation:

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Suppose you toss a fair coin 10 times, let X denote the number of heads. (a) What is the probability that X=5? (b) What is the p
zubka84 [21]

Answer:  The required answers are

(a) 0.25,    (b) 0.62,    (c) 6.

Step-by-step explanation:  Given that we toss a fair coin 10 times and X denote the number of heads.

We are to find

(a) the probability that X=5

(b) the probability that X greater or equal than 5

(c) the minimum value of a such that P(X ≤ a) > 0.8.

We know that the probability of getting r heads out of n tosses in a toss of coin is given by the formula of binomial distribution as follows :

P(X=r)=^nC_r\left(\dfrac{1}{2}\right)^r\left(\dfrac{1}{2}\right)^{n-r}.

(a) The probability of getting 5 heads is given by

P(X=5)\\\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}\\\\\\=\dfrac{10!}{5!(10-5)!}\dfrac{1}{2^{10}}\\\\\\=0.24609\\\\\sim0.25.

(b) The probability of getting 5 or more than 5 heads is

P(X\geq 5)\\\\=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}+^{10}C_6\left(\dfrac{1}{2}\right)^6\left(\dfrac{1}{2}\right)^{10-6}+^{10}C_7\left(\dfrac{1}{2}\right)^7\left(\dfrac{1}{2}\right)^{10-7}+^{10}C_8\left(\dfrac{1}{2}\right)^8\left(\dfrac{1}{2}\right)^{10-8}+^{10}C_9\left(\dfrac{1}{2}\right)^9\left(\dfrac{1}{2}\right)^{10-9}+^{10}C_{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{10-10}\\\\\\=0.24609+0.20507+0.11718+0.04394+0.0097+0.00097\\\\=0.62295\\\\\sim 0.62.

(c) Proceeding as in parts (a) and (b), we see that

if a = 10, then

P(X\leq 0)=0.00097,\\\\P(X\leq 1)=0.01067,\\\\P(X\leq 2)=0.05461,\\\\P(X\leq 3)=0.17179,\\\\P(X\leq 4)=0.37686,\\\\P(X\leq 5)=0.62295,\\\\P(X\leq 6)=0.82802.

Therefore, the minimum value of a is 6.

Hence, all the questions are answered.

3 0
3 years ago
3,969 + b^2 = 11,025
ludmilkaskok [199]

Answer:

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Step-by-step explanation:

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sqrt b^2 = sqrt 7056

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andreev551 [17]

Answer:

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144 = 16*H

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144/16 = 16/16 *H

9 = H

Now you have the length and height

Length: 16

Height: 9

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The top and bottom of the rectangle is the length which is 16 so

16 + 16 = 32

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9 + 9 = 18

Now add then both

32 + 18 = 50

So the perimeter of the rectangle is 50

Hope this helped!

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