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KiRa [710]
3 years ago
13

1. solve the following pair of simultaneous equations.

Mathematics
2 answers:
ruslelena [56]3 years ago
6 0
From equation 1 e=23-6f
sub 1 into 2. 2(23-6f)-f=7
46-12f-f=7
46-7-12f-f=0
39-11f=0
39=11f
then u divide both side by 11
therefore f=3.55
subtitude f in 1
6(3.55)+e=23
21.3+e=23
e=23-21.3
e=1.7
Arisa [49]3 years ago
4 0
1.7 is the correct answer.
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                42 minutes = (7/8 + 5/6 + x) / 1/18
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3x

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2 years ago
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nikdorinn [45]
Y = 3000x + 28000
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3 years ago
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Please help me , at least do one and explain it please .
Len [333]

9514 1404 393

Answer:

  1. 113.0 cm²
  2. 2464.0 in²
  3. 95.0 ft²

Step-by-step explanation:

1. The area of a circle is given by the formula ...

  A = πr² . . . . . where r is the radius

The radius is shown as 6 cm, so the area is ...

  A = (3.14)(6 cm)² = 113.0 cm²

__

2. The radius is shown as 28 in. We note that this is a number divisible by 7, so we choose 22/7 for π.

  A = (22/7)(28 in)² = 2464 in² . . . . see comment

__

3. The radius is half the diameter, so is 11/2 = 5.5 ft. Then the area is ...

  A = (3.14)(5.5 ft)² = 95.0 ft²

_____

<em>Additional comment</em>

If you use a more exact value of π for problem 2, the area is 2463.0 in². If you use 3.14 as the value of π, the area rounds to 2461.8 in². For these values of pi (3.14 or 22/7), the answer is only good to about 3 significant digits. 2464 has more significant digits, so digits beyond the first 3 may be in error.

7 0
2 years ago
40% of Oatypop cereal boxes contain a prize. Hannah plans to keep buying cereal until she gets a prize. What is the probability
aleksley [76]

Answer:

(0.6)^{n}+n(0.4)(0.6)^{n-1}+\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}

Step-by-step explanation:

We will use the binomial distribution. Let X be the random variable representing the no. of boxes Hannah buys before betting a prize.

Our success is winning the prize, p =40/100 = 0.4

Then failure q = 1-0.4  = 0.6

Hannah keeps buying cereal boxes until she gets a prize. Then n be no. times she buys the boxes.

P(X ≤ 3) = P(X=0) +P(X=1)+P(X=2)+P(X=3)

             = \binom{n}{0}p^{0}q^{n-0} +  \binom{n}{1}p^{1}q^{n-1}+ \binom{n}{2}p^{2}q^{n-2}+ \binom{n}{3}p^{3}q^{n-3}


             =  q^{n}+npq^{n-1}+\frac{n(n-1)(p)^{2}q^{n-2}}{2}+\frac{n(n-1)(n-2)p^{3}q^{n-3}}{6}

             = (0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}


7 0
3 years ago
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