It is desired to estimate the mean gpa of each undergraduate class at a large university. assume that the variance of the gpas is 1.44. how large a sample is necessary to estimate the mean gpa within 0.25 at the 99% confidence level

3x+7=22

3x=15

x=5

 the number is 5

Twenty-five squares with a side length of 6 inches are required to cover a square with a side length of 30 inches.

To find out how many squares with sides that are 6 inches long are needed to cover a square with a length of 30 inches, you first need to find the area of each square.

The area of a square is found by multiplying the length of one side by itself.

So, the area of the smaller square is 6 * 6 = 36 square inches, and

the area of the larger square is 30 * 30 = 900 square inches.

Then, you divide the area of the larger square by the area of the smaller square:

          900/36 = 25.

Therefore, 25 squares with sides of 6 inches are needed to cover a square with a side length of 30 inches.

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The solution to this system is (x, y) = (8, -22).

The y-values get closer together by 2 units for each 2-unit increase in x. The difference at x=2 is 6, so we expect the difference in y-values to be zero when we increase x by 6 (from 2 to 8).

You can extend each table after the same pattern.

In table 1, x-values increase by 2 and y-values decrease by 8.

In table 2, x-values increase by 2 and y-values decrease by 6.

The attachment shows the tables extended to x=10. We note that the y-values are the same (-22) for x=8 (as we predicted above). That means the solution is ...

... (x, y) = (8, -22)

Answer:

(8,-22)

Step-by-step explanation:

Table 1)

To form equation we will use two point slope form

[tex](x_1,y_1)=(-4,26)\\(x_2,y_2)=(-2,18)[/tex]

Formula :[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

Substitute the values :

[tex]y-26=\frac{18-26}{-2+4}(x+4)[/tex]

[tex]y-26=-4(x+4)[/tex]

[tex]y-26=-4x-16[/tex]

[tex]y=-4x-16+26[/tex]

[tex]y=-4x+10[/tex] ---1

Table 2)

To form equation we will use two point slope form

[tex](x_1,y_1)=(-4,14)\\(x_2,y_2)=(-2,8)[/tex]

Formula :[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

Substitute the values :

[tex]y-14=\frac{8-14}{-2+4}(x+4)[/tex]

[tex]y-14=-3(x+4)[/tex]

[tex]y-14=-3x-12[/tex]

[tex]y=-3x+2[/tex] ---2

Now we are supposed to solve 1 and 2

Substitute the value of y from 1 in 2

[tex]-4x+10=-3x+2[/tex]

[tex]8=x[/tex]

Substitute the value of x in 2

[tex]y=-3(8)+2[/tex]

[tex]y=-22[/tex]

Hence the solution to this system is (8,-22)

6y-8 =3y-2

subtract 3y from both sides

3y-8 = -2

add 8 to both sides

3y=6

y=6/3 =2

 y = 2

The given differential equation (3x + 6y) dx + (6x − 8y3) dy = 0 is exact because its mixed partial derivatives are equal. To solve it, first integrate M with respect to x to get a function involving an unknown function of y, then compare it with N to determine the unknown function.

To determine if the given differential equation is exact, we need to check if the partial derivative of M with respect to y is the same as the partial derivative of N with respect to x. In this given differential equation (3x + 6y) dx + (6x − 8y3) dy = 0, M = 3x + 6y and N = 6x - 8y3. First, compute the partial derivative of M with respect to y (∂M/∂y) which is 6, then the partial derivative of N with respect to x (∂N/∂x), which is 6 as well. Since ∂M/∂y=∂N/∂x, the given differential equation is exact.

To solve this exact differential equation, we integrate M dx, i.e., ∫M dx, to get ψ(x,y) =1.5x^2+6xy+h(y), where h(y) is an arbitrary function of y. Next, differentiate ψ(x,y) with respect to y, then compare the result with N: ψy=6x+dh/dy=6x-8y³. Solving for dh/dy gives dh/dy=-8y³, so integrating this gives h(y)=-2y⁴+C, where C is a constant. Therefore, the solution to the differential equation is ψ(x,y)=1.5x²+6xy-2y⁴=C.

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Answer is provided in the image attached.

Answer:[tex]\frac{5}{6}[/tex]

Step-by-step explanation:

Given

5 apples and 7 peaches out of which 4 apples and 2 peaches is rotten

i.e. 1 apple and 5 peaches is fresh

probability of fresh fruit is =[tex]\frac{6}{12}[/tex]

probability of apple =[tex]\frac{5}{12}[/tex]

Probability of both =[tex]\frac{1}{12}[/tex]

Probability of fresh or an apple is=[tex] \frac{6}{12}+\frac{5}{12}-\frac{1}{12}[/tex]

=[tex]\frac{5}{6}[/tex]

Answer:

The Meaning of term Variable is that Quantity Represented by small or Capital Alphabets of English whose value is not fixed.It can vary on different Situations.For example , Human diet can be called Variable.

In morning , you ate total of 0.25 Kg, in Afternoon you ate =0.50 Kg in total, and in the evening you ate total of 1 Kg.

So, if x is a variable of my Diet taken from Morning to evening, it has taken three distinct values, which are, x= 0.25, 0.50,1.

Here, x is a Variable, and 0.25,0.50 and 1 are Constant.

Answer:

The required equation is  [tex]y =\frac{(q-7)}{(p-6)}[/tex].

Step-by-step explanation:

Consider the provided equation.

[tex]py+7=6y+q[/tex]

We need to simplify the equation and convert it into a linear equation with unknown coefficients.

Subtract 6y from both sides.

[tex]py - 6y + 7 = q[/tex]

Subtract 7 from both sides.

[tex]py - 6y = q - 7[/tex]  

Take y common from left side.

[tex]y(p-6) = q-7[/tex]

 Divide both sides by p-6.

 [tex]y =\frac{(q-7)}{(p-6)}[/tex]

Hence, the required equation is  [tex]y =\frac{(q-7)}{(p-6)}[/tex].

There are 667 positive integers not exceeding 1000 that are not divisible by either 4 or 6.

This is a question about number theory and involves the comprehension of divisibility. To identify the positive integers that are not divisible by either 4 or 6 and not exceeding 1000 is an application of the principle of inclusion and exclusion.

First, we need to figure out how many numbers up to 1000 are divisible by 4, which is 1000/4 = 250. Then, we look at how many numbers are divisible by 6, which is 1000/6 = about 166 (we only consider the whole numbers).

However, some numbers are divisible by both 4 and 6, so we have over-counted and need to correct for this. These would be the numbers divisible by the least common multiple (LCM) of 4 and 6, which is 12. There are 1000/12 = about 83 such numbers.

Thus, according to the principle of inclusion and exclusion, the total number of numbers divisible by 4 or 6 would be 250 + 166 - 83 = 333. The last step is to subtract this from the total number of integers from 1 to 1000, so the answer would be 1000 - 333 = 667.

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All of the expressions are equivalent except (b) -4 + y.

All the expressions involve adding -4 and - y in various orders, resulting in equivalent expressions due to the commutative property of addition.

However, the expression -4 + y  is different because it combines the positive y term with the negative 4, resulting in - 4 + y.

The other expressions either have y added to -4 ( -4 - y) or have -4 added to y (-y - 4, -y + (-4)).

While addition is commutative, changing the order of the terms can affect the overall expression when dealing with negative terms.

So, while mathematically equivalent, the form of -4 + y stands out due to its structure.

Complete Question:

All of the following expressions are equivalent except _____.

(a) -4 - y

(b) -4 + y

(c) -y - 4

(d) -y + (-4 )

Answer:

66,000 units

Step-by-step explanation:

Break-Even Sales (units) = Fixed Costs ÷ Unit Contribution Margin = $561,000 ÷ ($8 + $0.50) = 66,000 units

The break-even point in units after the decrease in variable costs is 66,000 units.

Given that the fixed costs are $561,000 and the original unit contribution margin is $8.00, we first calculate the original break-even point:

[tex]\[ \text{Original Break-even point in units} = \frac{561,000}{8} \][/tex]

[tex]\[ \text{Original Break-even point in units} = 70,125 \text{ units} \][/tex]

Now, since variable costs are decreased by $0.50 a unit, the new unit contribution margin becomes:

[tex]\[ \text{New Unit Contribution Margin} = 8 + 0.50 \][/tex]

[tex]\[ \text{New Unit Contribution Margin} = 8.50 \][/tex]

Using the new unit contribution margin, we calculate the new break-even point:

[tex]\[ \text{New Break-even point in units} = \frac{561,000}{8.50} \][/tex]

[tex]\[ \text{New Break-even point in units} = 66,000 \text{ units} \][/tex]

Therefore, the new break-even point in units, after the decrease in variable costs, is 66,000 units.