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Kisachek [45]
2 years ago
10

A contractor needs to buy nails to build a house. The nails come in small boxes and large boxes. Each small box has 100 nails an

d each large box has 350 nails. The contractor bought a total of 9 boxes that have 2400 nails altogether. Write a system of equations that could be used to determine the number of small boxes purchased and the number of large boxes purchased. Define the variables that you use to write the system.
Mathematics
1 answer:
IRISSAK [1]2 years ago
6 0

Answer:

3 small boxes and 6 large ones

Step-by-step explanation:

3 times 100 equals 300 and 6 times 350 is 2100 and 2100 plus 300 is 2400.

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Caleb solved this equation and recorded his work. 7.4x + 4.1(2x − 4) = −2.3(x − 6) − 21.6 1. 7.4x + 8.2x − 16.4 = −2.3x + 13.8 −
IRINA_888 [86]
For this case we have the following equation:
 7.4x + 4.1 (2x - 4) = -2.3 (x - 6) - 21.6

 Step 1 of the solution is to make the distributive property.
 We have then:
 7.4x + 8.2x - 16.4 = -2.3x + 13.8 - 21.6

 Step 1 is correct.
 The step is to combine similar terms. Caleb wrote: 7.4x + 8.2x - 16.4 = -2.3x + 35.4

 He wrongly combined the similar terms on the right side of the equation.
 The correct solution is: 7.4x + 8.2x - 16.4 = -2.3x -7.8

 Answer:
 In step 2, the like terms were not combined on the right side of the equation.
6 0
2 years ago
Read 2 more answers
An airline, believing that 5% of passengers fail to show for flights, overbooks (sells more tickets than there are seats). Suppo
Fittoniya [83]

Answer:

Q1. 13 passengers

Q2. 0.1756 (approx. 0.18)

Step-by-step explanation:

Q1. 267 seats are available on the plane

5% is expected to fail to show up

Hence, no of passengers expected not to show up = 267 * 0.05

= 13.35 (approx 13 passengers)

Q2. See working in the attachment as I had to explain it step by step.

6 0
2 years ago
In a statewide soccer competition, 36 games were played. Each team participating in the competition played once with every other
Tomtit [17]

Answer:

9 teams

Step-by-step explanation:

If the total games played was 36 and no team played each other twice, we need to ensure there isn't any double counting.

36 = (n-1) + (n-2) + (n-3) ... + (n-(n-1))

using this knowledge, we can then count up:

1+2+3+4+5+6+7+8 = 36

If our highest number is 8, then we know there must be 9 teams, because no team can play themselves.

8 0
2 years ago
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se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y
alexandr1967 [171]

Answer:

It is proved that f_x, f_y exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)

  • To show exixtance of f_x(0,0), f_y(0,0) we take,

f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0    exists.

And,

f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}   exists.

  • To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}  where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

4 0
2 years ago
At a financial institution, a fraud detection system identifies suspicious transactions and sends them to a specialist for revie
labwork [276]

Answer:

a. E(X) = 54.4

b. E(X) = 2.5

c. P(Y=2) = .0116

Step-by-step explanation:

a.

    E(X) = np = .40 probability * 136 trials = 54.4 blocked transmissions

    To get the expected value, we simply multiply probability times number of trials. You can look at it in simple terms by thinking if there's a 50% chance of flipping heads and you flip a coin twice, in an ideal world you will have .5*2 = 1 head.

b.

    i. Let X represent the number of suspicious transmissions reviewed until finding the first blocked one. We will use a geometric distribution to model the "first" transmission. Whenever we're looking for the "first" time something happens, we use geometric.

   ii. E(X) = 1/p , according to the geometric model.

              = 1/.4 = 2.5.

       We expect that the specialist will review 2.5 suspicious transactions <em>on average </em>before finding the first transmission that will be blocked.

c.

    i. Let Y represent the exact number of blocked transmissions out of 10. We will use a binomial distribution to model the "fixed" number of transmissions. Whenever we're looking for a "fixed" number of times something happens, we use binomial.

    ii. P(Y=k) = (n choose k)(p^k)(q^n-k)

        P(Y=2) = (¹⁰₂)(.4^2)(.6^10-2)

                    = 45 (.4^2)(.6^10-2) = .0016

        As for calculator notation, the n choose k can be accessed on a TI-84 via MATH -> PRB -> nCr. It looks like 10 nCr 2 on the display.

        Hence the probability that two transactions out of ten will be blocked is .0016 by the binomial model.

5 0
2 years ago
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