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2 years ago

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A school administrator believes that the mean class gpas for a given course are higher than the preferred mean of 2.75 at a sign

ificance level of 0.05 , and checks a randomly chosen sample of 50 classes. create a histogram, and calculate , the -statistic, and the -value.

1 answer:

Svetach [21]2 years ago

4 0

The mean GPA is greater than 2.75.

<h3>What is GPA?</h3>

A grade point average is a figure that represents the average of the final grades obtained in courses over time.
A student's grade point average, sometimes known as a GPA, is computed by adding all accumulated final grades and dividing the total by the number of grades granted.

To find the GPA:

GPA is calculated and could be utilized by the college to tell choices, which include the subsequent educational progression.
Divide the total range of grade factors earned with the aid of the total number of letter-graded units undertaken.
Mean GPA = total/50
= 2.75.

Therefore, the mean GPA is greater than 2.75.

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Find the formula for the area of the shaded region in the figure.<br> x² = 4py<br> ہے

Alexandra [31]

The formula for the area of the shaded region in the figure.

x² = 4py is mathematically given as

$A=\frac{8}{3} \sqrt{p h} \cdot h$

<h3>What is the formula for the area of the shaded region in the figure?</h3>

Parameters

$x^{2}=4 p y$

$y=h$

$\therefore x^{2}=4 p h \Rightarrow x=2 \sqrt{p h} \\$

Generally, the equation for the Area is mathematically given as

$& A=2 \int_{0}^{2 \sqrt{p h}}\left(h-\frac{x^{2}}{4 p}\right) d x \\\\& A=2\left(h x-\frac{x^{3}}{12 p}\right]_{0}^{2 \sqrt{p h}} \\\\& A=2\left[\left[h 2 \sqrt{p h}-\frac{1}{12 p} \cdot(2 \sqrt{p h})^{3}-0\right]\right. \\$

$&A=2\left[2 h^{3 / 2} \sqrt{p}-\frac{8(\sqrt{p})^{3}(\sqrt{h})^{3}}{12 p}-0\right] \\\\&A=2\left[2 h^{k / 2} \sqrt{p}-\frac{2}{3} \sqrt{p} \cdot h^{3 / 2}\right] \\\\&A=2 \times \frac{4 \sqrt{p} \cdot h^{3 / 2}}{3} \\\\&A=\frac{8}{3} \cdot \sqrt{p} \cdot \sqrt{h} \cdot h \\\\&A=\frac{8}{3} \sqrt{p h} \cdot h$

$A=\frac{8}{3} \sqrt{p h} \cdot h$

In conclusion, the equation for the Area is

$A=\frac{8}{3} \sqrt{p h} \cdot h$

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2 years ago

Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie

melamori03 [73]

Answer:

The complete solution is

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

$y = e^{mx}$

Differentiating with respect to x

$y'= me^{mx}$

Again differentiating with respect to x

$y''= m ^2 e^{mx}$

Putting the value of y, y' and y'' in left side of the differential equation

$3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0$

$\Rightarrow 3m^2-8m-3=0$

The auxiliary equation is

$3m^2-8m-3=0$

$\Rightarrow 3m^2 -9m+m-3m=0$

$\Rightarrow 3m(m-3)+1(m-3)=0$

$\Rightarrow (3m+1)(m-3)=0$

$\Rightarrow m = 3, -\frac13$

The complementary function is

$y= Ae^{3x}+Be^{-\frac13 x}$

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)

For particular integral

$y_p=\frac 1{(3D+1)(D-3)} .4$

$=4.\frac 1{(3D+1)(D-3)} .e^{0.x}$ [since $e^{0.x}=1$ ]

$=4\frac{1}{(3.0+1)(0-3)}$ [ replace D by 0 , since L(0)≠0]

$=-\frac43$

The complete solution is

y= C.F+P.I

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

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3 years ago

Find the lateral area of the cylinder.<br><br>Give your answer in the terms of pi.

Wewaii [24]

Hi there!

The formula for the lateral area of a cylinder is LA = 2 x pi x r x h. (two times pi times radius times height) Using this formula, we can plug in the values and solve for the lateral area.

Plugging in the values: LA = 2 x pi x 7 x 9
Simplifying: LA = 2pi x 63
LA = 126pi yd^2

ANSWER:
The 4th option - 126pi yd^2

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

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3 years ago

What's the sum of 3/8 and 1/16? a. 1/6 b. 7/16 c. 4/24 d. 1/4

Norma-Jean [14]

Answer: B

Step-by-step explanation: 3/8 + 1/16 = 0.4375 and 7/16 = 0.4375

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MrRa [10]

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3 years ago

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