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Kamila [148]
3 years ago
8

Simplify your expression 3.7-5.3n-4n+5 can somebody help?

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
5 0

Answer:

-9.3n + 8.7

Step-by-step explanation:

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L is the midpoint of line segment KM. If KL = 2x - 3 and LM = x + 7. find the value of x.
tatyana61 [14]

Answer:

x = 10

Step-by-step explanation:

Since L is the midpoint

KL = LM

2x-3 = x+7

Subtract x from each side

2x-3-x = x+7-x

x-3 = 7

Add 3 to each side

x-3+3 = 7+3

x = 10

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3 years ago
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Are x and y inversely proportional in the table below?
rosijanka [135]
Answer: yes they are

Example: if you don’t believe me check for yourself :)
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2 years ago
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Substitution for:<br> X-5y=10<br> 2x-10y=20
sergey [27]
Since you need an isolated variable to use the substitution method, we need to re-arrange one of the equations. This will probably be easiest to do with the first one.
Add 5y to both sides of the first equation.
x=10+5y
Now, in the second equation, put in 10+5y in any spot that has an x.
2(10+5y)-10y=20
Distribute the 2 to both numbers in the parenthesis.
20+10y-10y=20
Combine like terms.
20=20
This means that the two equations are actually the same. You can see this if you multiply the whole first equation by 2
2(x-5y=10)
2x-10y=20, which is the same as the second equation. Therefore, the two equations are actually the same one.
5 0
3 years ago
PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST
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

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

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)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

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