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KATRIN_1 [288]
2 years ago
9

Helpppp plsss!!!!!!

Mathematics
2 answers:
Vanyuwa [196]2 years ago
5 0
A because -1/4x is your slope and the +5 would just be shifted over to the 5 on the y-axis
Katen [24]2 years ago
3 0

Answer: y = -1/4x +5 (option A)

Step-by-step explanation:

you should substitute the point in the equation so, you have to use "+c" instead of the "-3". the equation you have to use is y = -1/4x + c

(4,4)

y = -1/4x + c

4 = -1/4(4) + c

4 = -1 + c

4+1=c

c = 5

So the -3 in the equation should be replaced with 5 in the equation and that's the parallel line passing through the point (4,4)

therefore the equation is -   y = -1/4x +5

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62

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Researchers fed mice a specific amount of Dieldrin, a poisonous pesticide, and studied their nervous systems to find out why Die
Elodia [21]

Answer:

Step-by-step explanation:

Part A

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.

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3 years ago
Answer to this??? Need help
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A) 2 and 7
B) 2 and 6
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