Y = -4× - 5y = -4× + 5
1 answer:
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Answer:
7.41 m/s
Step-by-step explanation:
Force of a car inclined at an angle
F=mg Sin Θ where; m = 2956 kg , g = 9.8m/s² , Θ = 35⁰
F = 2956 x 9.8 x Sin 35 = 16615.82 N
Net force due to friction = 16615.82 - 3080 = 13535.82 N
To calculate acceleration; F= Ma
13535.82 = 2956 x a
a = 4.58 m/s²
From equation of motion
V² = 0 + (2 x 4.58 x 6)
V =
V = 7.41 m/s
Answer:
The 90% confidence interval for the difference in the mean grade on test 7 for all students who attended a review session and for all students who did not attend a review session is (3.7, 12.3).
Step-by-step explanation:
Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
Subtraction between normal variables:
When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.
A simple random sample of 22 students who attended a review session was selected, and the mean grade on test 7 for this sample of 22 students was 85 with a standard deviation of 9.8.
This means that:
An independent simple random sample of 47 students who did not attend a review session was selected, and the mean grade on test 7 was 77 with a standard deviation of 10.6.
This means that:
Distribution of the difference in the mean grade on test 7 for all students who attended a review session and for all students who did not attend a review session.
Mean:
Standard deviation:
Confidence interval:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
That is z with a pvalue of , so Z = 1.645.
Now, find the margin of error M as such
So
The lower end of the interval is the sample mean subtracted by M. So it is 8 - 4.3 = 3.7
The upper end of the interval is the sample mean added to M. So it is 8 + 4.3 = 12.3
The 90% confidence interval for the difference in the mean grade on test 7 for all students who attended a review session and for all students who did not attend a review session is (3.7, 12.3).