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mihalych1998 [28]
3 years ago
5

Julio tested the point (4, –2) to see whether it is a solution to this system of equations. –3x – 2y = –8, y = 2x – 5 His work i

s shown below.
Mathematics
2 answers:
Vaselesa [24]3 years ago
7 0

The answer is B. No he did not check the point in the second equation.

kolezko [41]3 years ago
6 0

if it is solution point (4, –2) must satisfy both equations
(1) –3x – 2y = –8-->-3*4-2*(-2)=-8
-12+4=-8 then-->-8=-8   ok
(2)  y = 2x – 5
-2=2*4-5=3---------->-2 is not 3

the point  (4, –2) is not solution of the system of equations
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Solve the equation.
Levart [38]

Answer:

x=34

Step-by-step explanation:

6 - ( x-7) ^ 1/3 = 3

Subtract 6 from each side

6-6 - ( x-7) ^ 1/3 = 3-6

- ( x-7) ^ 1/3 = -3

Divide each side by a negative

 ( x-7) ^ 1/3 = 3

Cube each side

 ( x-7) ^ 1/3 ^3 = (3)^3

x-7 = 27

Add 7 to each side

x-7+7 = 27+7

x = 34

Check

6 - ( 34-7) ^ 1/3 = 3

6 - (27^1/3 = 3

6 -3 =3

3=3

Good solution

8 0
3 years ago
Can someone help me please
soldier1979 [14.2K]

Step-by-step explanation:

c >= 2

that means any value of c greater or equal to 2 is a valid solution. so, yes, 2 is a solution for this.

c < 2

that means any value of c smaller than 2 is a valid solution. so, no, 2 is not smaller than 2, so it is not a solution.

c < 3

that means any value of c smaller than 3 is a valid solution. so, yes, 2 is smaller than 3, so 2 is a solution.

3 < c

that means any value of c, for that 3 is smaller, is a valid solution. our in other words, any value for c larger than 3 is a valid solution. so, no, 2 is not larger than 3, so it is not a solution.

-8 < c

that means any value of c, for that -8 is smaller, is a valid solution. our in other words, any value for c larger than -8 is a valid solution. so, yes, 2 is larger than -8, so it is a solution.

5 0
2 years ago
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7 0
3 years ago
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7 0
3 years ago
For each of the following scenarios state whether H0 should be rejected or not. State any assumptions that you make beyond the i
scoundrel [369]

Answer:

a)H_0 :\mu = 4\\ H_1 : \mu \neq 4 , n = 15 , X=3.4 , S=1.5 , α = .05

Formula : t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}

t = \frac{3.4-4}{\frac{1.5}{\sqrt{15}}}

t =-1.549

p- value = 0.607(using calculator)

α = .05

p- value > α

So, we failed to reject null hypothesis

b)H_0 :\mu = 21\\ H_1 : \mu < 21 , n =75 , X=20.12 , S=2.1 , α = .10

Formula : t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}

t = \frac{20.12-21}{\frac{2.1}{\sqrt{75}}}

t =-3.6290

p- value = 0.000412(using calculator)

α = .1

p- value< α

So, we reject null hypothesis

(c) H_0 :\mu = 10\\ H_1 : \mu \neq 10, n = 36, p-value = 0.061.

Assume α = .05

p-value = 0.061.

p- value > α

So, we failed to reject null hypothesis

7 0
3 years ago
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