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Vsevolod [243]
3 years ago
5

One common system for computing a grade point average​ (GPA) assigns 4 points to an​ A, 3 points to a​ B, 2 points to a​ C, 1 po

int to a​ D, and 0 points to an F. What is the GPA of a student who gets an A in a
33​-credit
​course, a B in each of
three

2​-credit
​courses, a C in a
3​-credit
​course, and a D in a
2​-credit
​course?
Mathematics
1 answer:
Dafna1 [17]3 years ago
6 0

GPA = 3.59

For calculating the GPA, first calculate the quality points for each grade and course and then add them up like shown below:

A - 4 points x 33 credit hours = 132 quality points

B - ( 3 points x 2 credit hours ) x 3 (because there are 3 courses) = 18 quality points

C - 2 points x 3 credits = 6 quality points

D - 1 point x 2 credits = 2 quality points

Now add up all the quality points = 132 + 18 + 6 + 2 = 158

Add the credit hours as well = 33 + 6 + 3 + 2 = 44

Divide the total quality points by total credit hours to get the GPA = 158/44 = 3.59

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Answer:

We conclude that the board's length is equal to 2564.0 millimeters.

Step-by-step explanation:

We are given that a sample of 26 is made, and it is found that they have a mean of 2559.5 millimeters with a standard deviation of 15.0.

Let \mu = <u><em>population mean length of the board</em></u>.

So, Null Hypothesis, H_0 : \mu = 2564.0 millimeters    {means that the board's length is equal to 2564.0 millimeters}

Alternate Hypothesis, H_A : \mu \neq 2564.0 millimeters      {means that the boards are either too long or too short}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                             T.S.  =  \frac{\bar X-\mu}{\frac{s }{\sqrt{n}} }  ~  t_n_-_1

where, \bar X = sample mean length of boards = 2559.5 millimeters

            s = sample standard deviation = 15.0 millimeters

             n = sample of boards = 26

So, <em><u>the test statistics</u></em> =  \frac{2559.5-2564.0}{\frac{15.0 }{\sqrt{26}} }  ~   t_2_5

                                     =  -1.529    

The value of t-test statistics is -1.529.

Now, at a 0.05 level of significance, the t table gives a critical value of -2.06 and 2.06 at 25 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the board's length is equal to 2564.0 millimeters.

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Given y = log3(x + 4), what is the range?​
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You can take the log of a negative or 0 number.

So x+4>0 implies x>-4.  (I just subtract 4 on both sides.)

So the domain is x>-4. You should see this also when you graph the curve that the curve only exist to the right of -4.

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If we look at the domain and range of this we can just swap it to get the domain and range of y=\log_3(x+4).

y=3^x-4 is an exponential function of 3^x that has been moved down 4 units.

The range since it has been moved down 4 units is (-4,\infty).

The domain of an exponential function is all real numbers.  There are no restrictions on what you can plug in for x.

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Range : (-\infty,\infty)

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