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Lemur [1.5K]
3 years ago
11

Answers?? With work please

Mathematics
1 answer:
Naya [18.7K]3 years ago
7 0
1.)-2
2)-2
3.)-3
4.)-1
5)1
6.)-6
7.)-6
8.)4
9.)0
10.)no solution
11.)-3
12.)all real numbers
13.)-1
14.)5
On working on the work for the first problem so u understand
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Choose the correct x-intercepts for the following quadratic fiction
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Could you please add a photo or the quadratic function?

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The voltage in a circuit is the product of two factors, the resistance and the current. If the voltage is 6ir + 15i + 8r+20, fin
andreev551 [17]

Answer:

  • resistance: (2r +5)
  • current: (3i +4)

Step-by-step explanation:

The factors of the given expression are ...

  6ir +15i +8r +20 = (3i +4)(2r +5)

Which factor is current and which is resistance is not clear. Usually, resistance is referred to using the variable r, so we suppose the expressions are supposed to be ...

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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
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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
What is the nearest degree ?
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Answer:

maybe 65 degrees

sorry if its wrong

6 0
3 years ago
Read 2 more answers
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