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sergij07 [2.7K]
3 years ago
5

64y-4+2y=-2 solve the equation

Mathematics
2 answers:
-Dominant- [34]3 years ago
6 0

Answer:33=y

Step 64u to 2y you get 66y the. -2+4 and you get 2 which is 66y=2. Now divide 66 by two to get 33=y

givi [52]3 years ago
4 0
First step is to isolate the variable.
Anything that doesn’t have a ‘y’ is have to be moved which means the -4 will change to positive 4 and added to both sides.
64y -4 +2y = -2
(64y+2y) -4 = -2
+4 +4
Which crosses out the 4s on the left side and leaves 66y = 2.
Divide 66 on both sides.
66y/66 = 2/66
(2/66) / 2 = 1/33

Answer: y= 1/33
I hope this helps..

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Select all equations that have two solutions. a x^2=16 b 4x^2=0 c x^2=-16 d 3x+2=14 e x^2-1=24 f (x+8)(x-8)=0
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5 0
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ohaa [14]

Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

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Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

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x = sample mean = 2.48

µ = population mean = 2.3

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t = (2.48 - 2.3)/(0.16/√6) = 2.76

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Step-by-step explanation:

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