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Ad libitum [116K]
3 years ago
12

Solve -8 = 2y – 2. Show your work and how to check your solution.

Mathematics
1 answer:
Mashcka [7]3 years ago
4 0
Answer: y = -3

Explanation:

-8 = 2y - 2
-2y = -2 + 8
y = 6/-2
y = -3
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Solve this system of equations: y = x2 – 3x + 12 y = –2x + 14 1.Isolate one variable in the system of equations, if needed. y =
lina2011 [118]
-2x+14=x²-3x+12
add 2x on both side:
14=x²-x+12
subtract 14 on both side:
x²-x-2=0
factor: (x-2)(x+1)=0
x=2 or x=-1
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8 0
3 years ago
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Alex has collected apples. It has 40 boxes of 8.5 kg each and 6 sacks of 90kg each. Pack the apples in bags of 2.5 kg each. how
zimovet [89]

Answer:

40*8.5=340

6*90=540

total=880/2.5=352bags

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3 years ago
Plese HELP !!!
beks73 [17]

Answer:

-9/8

Step-by-step explanation:

\frac{-2a^{-3}b^2}{a} \\ \\ =-\frac{2b^2}{a^4} \\ \\ =-\frac{2(3)^2}{(-2)^4} \\ \\ =-\frac{18}{16} \\ \\ =-\frac{9}{8}

6 0
1 year ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

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

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
3 years ago
Which set of equations is enough information to prove that lines c and d are parallel lines cut by transversal p?
ikadub [295]

Answer:

56

Step-by-step explanation:

4 0
3 years ago
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