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liraira [26]
4 years ago
14

Y=1/2x +5 Y=-3/2x-7 Find x and y

Mathematics
1 answer:
gayaneshka [121]4 years ago
8 0
I'm going to use the substitution method. 

If y = - 3/2 - 7, then:

1/2x + 5 = - 3/2x - 7

Combine like terms: 

4/2x = - 12. Mutiply both sides by 2/4 to get x = -12 (2/4)

Simplify to get:

x = - 6.

Plug - 6 back in for x in either equation.

Y = 1/2( - 6) + 5 which becomes Y = - 3 + 5.

X = - 6, Y = 2





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Answer:

29 ft x 58 ft

Step-by-step explanation:

Let x be the length of each side perpendicular to the wall, and y be the length of the side parallel to the wall.

The amount of wire available is:

116 = 2x+y\\y=116-2x

The area of the region is:

A=xy=x(116-2x)\\A(x)=116x-2x^2

The value of 'x' for which the derivate of the area function is zero will yield the maximum area:

A(x)=116x-2x^2\\A'(x) = 116-4x=0\\x=29\ ft

The value of y is:

y=116-2*29\\y=58\ ft

The dimensions of the region with the largest area are 29 ft x 58 ft.

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3 years ago
Clairissa is considering two job offers. One has an annual sallary of $61.1K
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Answer:

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Step-by-step explanation:

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$61.1K will give a weekly salary of $1.17K

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7 0
3 years ago
Read 2 more answers
According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones
Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = <u><em>back-to-college family spending on electronics</em></u>

SO, X ~ Normal(\mu=237,\sigma^{2} =54^{2})

The z score probability distribution for normal distribution is given by;

                                 Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean family spending = $237

           \sigma = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

        P(X < $150) = P( \frac{X-\mu}{\sigma} < \frac{150-237}{54} ) = P(Z < -1.61) = 1 - P(Z \leq 1.61)

                                                             = 1 - 0.9463 = <u>0.0537</u>

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

        P(X > $390) = P( \frac{X-\mu}{\sigma} > \frac{390-237}{54} ) = P(Z > 2.83) = 1 - P(Z \leq 2.83)

                                                             = 1 - 0.9977 = <u>0.0023</u>

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

     P($120 < X < $175) = P(X < $175) - P(X \leq $120)

     P(X < $175) = P( \frac{X-\mu}{\sigma} < \frac{175-237}{54} ) = P(Z < -1.15) = 1 - P(Z \leq 1.15)

                                                         = 1 - 0.8749 = 0.1251

     P(X < $120) = P( \frac{X-\mu}{\sigma} < \frac{120-237}{54} ) = P(Z < -2.17) = 1 - P(Z \leq 2.17)

                                                         = 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>

5 0
4 years ago
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