The function f(x) = –x2 + 60x – 116 models the monthly profit, in dollars, a shop makes for repairing windshields, where x is the number of windshields repaired, and f(x) is the amount of profit. Part A: Determine the vertex. What does this calculation mean in the context of the problem? (5 points) Part B: Determine the x-intercepts. What do these values mean in the context of the problem? (5 points)

need to kind K

Y=k x 1/x^2 = k/x^2

Find K when y=4 & x=5

Y=k/x^2 =

4=k/5^2=

4=k/25

K=4*25 =100

When x = 2

Y=100/2^2

Y=100/4

Y=25

the equivalent equation of the equation 4s=t+2 is s = [tex]\frac{t+2}{4}[/tex] .

Equivalent equations are algebraic equations that have identical solutions or roots. Adding or subtracting the same number or expression to both sides of an equation produces an equivalent equation. Multiplying or dividing both sides of an equation by the same non-zero number produces an equivalent equation.

According to the question

The equation  

4s=t+2

Now,

its Equivalent equations  is  :

Dividing equation by 4 both side  

i.e

s = [tex]\frac{t+2}{4}[/tex]

Hence , the equivalent equation of the equation is s = [tex]\frac{t+2}{4}[/tex] .

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The test result is not significant at α = 0.05 or α = 0.01.

For a hypothesis test of H0:p1 − p2 = 0 against the alternative Ha:p1 − p2 ≠ 0, the test statistic is found to be 2.2. Given the information provided, the result is not significant at either α = 0.05 or α = 0.01. This conclusion is drawn based on the p-value and the 95% confidence interval.

On compared the test statistic to critical values for α = 0.05 and α = 0.01. Since the test statistic falls within the critical region for α = 0.05 but not for α = 0.01, we concluded that the result is significant at α = 0.05 but not at α = 0.01. The correct options B.

To determine the significance of the test statistic at different levels of significance, we need to compare it to critical values associated with the chosen alpha levels.

Given that the test statistic is 2.2, we need to refer to the critical values of the test statistic for a two-tailed test at α = 0.05 and α = 0.01. These critical values are typically obtained from statistical tables or software.

Let's assume the critical value at α = 0.05 is [tex]\( z_{\alpha/2} = \pm 1.96 \)[/tex]and the critical value at α = 0.01 is [tex]\( z_{\alpha/2} = \pm 2.58 \)[/tex](for a standard normal distribution).

If the test statistic falls within the range defined by these critical values, we can conclude that the result is significant at the corresponding alpha level. Otherwise, the result is not significant.

Since the test statistic of 2.2 falls between the critical values of [tex]\( \pm 1.96 \)[/tex] for α = 0.05 but outside the critical values of [tex]\( \pm 2.58 \)[/tex] for α = 0.01, we can conclude that:

The result is significant at α = 0.05 but not at α = 0.01.

Final answer: The result is significant at α = 0.05 but not at α = 0.01.

We compared the test statistic to critical values for α = 0.05 and α = 0.01. Since the test statistic falls within the critical region for α = 0.05 but not for α = 0.01, we concluded that the result is significant at α = 0.05 but not at α = 0.01. This interpretation aligns with standard hypothesis testing procedures.

Complete question

For a hypothesis test of H0:p1 − p2 = 0 against the alternative Ha:p1 − p2 ≠ 0, the test statistic is found to be 2.2. Which of the following statements can you make about this finding?

A)The result is significant at both α = 0.05 and α = 0.01.

B)The result is significant at α = 0.05 but not at α = 0.01.

C)The result is significant at α = 0.01 but not at α = 0.05.

D)The result is not significant at either α = 0.05 or α = 0.01.

E)The result is inconclusive because we don't know the value of p.

Answer:

60 miles.

Step-by-step explanation:

We have been given that the scale map is 0.5 inch : 20 miles. On the map, the distance between two cities is 1.5 inches. We are asked to find the actual distance between the two cities.

We will use proportions to solve for the actual distance between both cities as:

[tex]\frac{\text{Actual distance}}{\text{Map distance}}=\frac{20\text{ miles}}{\text{0.5 inch}}[/tex]

[tex]\frac{\text{Actual distance}}{\text{1.5 inches}}=\frac{20\text{ miles}}{\text{0.5 inch}}[/tex]

[tex]\frac{\text{Actual distance}}{\text{1.5 inches}}*\text{1.5 inches}=\frac{20\text{ miles}}{\text{0.5 inch}}*\text{1.5 inches}[/tex]

[tex]\text{Actual distance}=20\text{ miles}*3[/tex]

[tex]\text{Actual distance}=60\text{ miles}[/tex]

Therefore, the actual distance between both cities is 60 miles.

The largest number of goody bags that Kara can make are [tex]18[/tex] .

Highest or greatest Common Factor is the largest common factor that all the numbers have in common.

We have,

Number of Lollipops [tex]=90[/tex]

Number of chocolate bars [tex]=36[/tex]

Number of gumballs [tex]=72[/tex]

So,

To find the number of bags;

First find out the Highest Common Factor of [tex]90,36,72[/tex];

[tex]90=2*3*3*5[/tex]

[tex]36=2*2*3*3[/tex]

[tex]72=2*2*2*3*3[/tex]

So, from the factors of all numbers we have,

Highest Common Factor [tex]=18[/tex]

Now,

Lollipops [tex]=\frac{90}{18} =5[/tex]

Chocolate bars [tex]=\frac{36}{18} =2[/tex]

Gumballs [tex]=\frac{72}{18} =4[/tex]

So, the largest number of goody bags that Kara can make are [tex]18[/tex] so that each goody bag has [tex]5[/tex] number of lollipops, [tex]2[/tex] number of chocolate bars, and [tex]4[/tex] number of gumballs.

Hence, we can say that the largest number of goody bags that Kara can make are [tex]18[/tex] .

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Answer:

There are two solutions [tex]x=\frac{5}{2},-\frac{5}{2}[/tex]

Step-by-step explanation:

Given : Equation [tex]16x^2=100[/tex]

To find : How many solutions will there be to the following equation?      

Solution :

Equation [tex]16x^2=100[/tex]

Solve the equation,

Divide by 16 both side,

[tex]\frac{16x^2}{16}=\frac{100}{16}[/tex]

[tex]x^2=\frac{100}{16}[/tex]

Taking root both side,

[tex]x=\sqrt{\frac{100}{16}}[/tex]

[tex]x=\sqrt{\frac{10^2}{4^2}}[/tex]

[tex]x=\pm\frac{10}{4}}[/tex]

[tex]x=\pm\frac{5}{2}}[/tex]

Therefore, There are two solutions [tex]x=\frac{5}{2},-\frac{5}{2}[/tex]

The area of smaller triangle is 59 square feet

Two triangles are similar if they have the same ratio of corresponding sides and equal pair of corresponding angles.

The formula for the area of triangle is

[tex]Area = \frac{1}{2} \times base \times \ height[/tex]

According to the given question.

The area of the larger triangle is 105 square feet.

And the one side of the larger triangle and the smaller traingle is 16 and 12 feet respectively.

Suppose the height of thesmaller triangle be x feet and the height of the larger triangle be y feet.

Since, the corresponding edges of similar triangles are proportional.

Therefore,

[tex]\frac{y}{x} = \frac{16}{12}[/tex]

[tex]\implies y = \frac{4}{3} x[/tex]

Also, the area of larger triangle is 105 square feet.

[tex]\implies \frac{1}{2} \times \frac{4}{3} x \times 16 = 105[/tex]

The above euqtaion can be written as

[tex]\implies \frac{1}{2}\times \frac{4}{3} x \times \frac{4}{3} \times 12 = 105[/tex]

[tex]\implies \frac{1}{2} \times (\frac{4}{3} )^{2} \times x \times 12 = 105[/tex]

[tex]\implies \frac{1}{2} \times x \times 12 = 105 \times \frac{9}{16}[/tex]

[tex]\implies \frac{1}{2} \times x \times 12 = 59[/tex]

⇒ Area of smaller triangle  = 59 square feet

Hence, the area of smaller triangle is 59 square feet.

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Answer:

Area will increase by 4 times

Step-by-step explanation:

Given: A rectangle has a base of 3 inches and a height of 9 inches.  

To find: If the dimensions are doubled, what will happen to the area of the rectangle?

Solution:

It is given that a rectangle has a base of 3 inches and a height of 9 inches.

Now, to find if the dimensions are doubled what will happen to the area, first we need to find the original area.

Original area of rectangle, when base is 3 inches and height 9 inches is

[tex]9\times3=27[/tex] square inches

Now, when dimensions are doubled , the base becomes 6 inches and height becomes 18 inches

So, new area becomes [tex]6\times18=108[/tex] square inches.

Now,

[tex]\frac{\text{new area}}{\text{original area} }=\frac{108}{27}[/tex]

[tex]\implies\frac{\text{new area}}{\text{original area} }=\frac{4}{1}[/tex]

Hence, the area will increase by 4 times.

Answer:

x^2/25 + y^2/81 =1

Step-by-step explanation:

solved it

Answer: c.  

6  

4

Step-by-step explanation:

The exponential function is given by :-

[tex]y=Ab^x[/tex], where A is the initial amount , b is common ratio and x is time period.

We know that in exponential functions, the ratio of the consecutive value of y is same.

From the table ,  [tex]b=\dfrac{6}{4}=\dfrac{3}{2}[/tex]

At x = 0

[tex]9=A(\dfrac{2}{3})^0\\\\\Rightarrow\ A=9[/tex]

At x=1

[tex]y=9(\dfrac{2}{3})^1=6[/tex]

At x=2

[tex]y=9(\dfrac{2}{3})^2=4[/tex]

Hence, the missing values for the exponential function : c.  6  4.

The side of the triangle which is the hypotenuse side is [tex]2\sqrt{13}[/tex]

Using trigonometry, Trigonometry is the area of mathematics that deals with particular angle functions and how to use them in computations. In trigonometry, an angle can have six common functions. The terms sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc) are their names and abbreviations.

[tex]c^{2} =4^{2} +6^{2} \\\\c^{2}=16+36\\\\c^{2}=52\\\\c=\sqrt{52} \\\\c= 2\sqrt{13}[/tex]

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Answer:

Option a) They are adjacent angles

Step-by-step explanation:

Given is a picture of a graph with angles 1 to 6 around it.

Out of these angle 2 and 5 are right angles.

1 and 2 are adjacent to each other.

They are not complementary because 1+2 not equals 90

They are neither supplementary since sum does not equal 180

They cannot be vertical because they are not formed by intersection of two lines.

Hence only option a is right

Option a) They are adjacent angles

The minimum value of the function occurs at [tex]\( x = -\frac{5}{2}[/tex], and the function reaches its minimum value of [tex]\( -\frac{1}{4} \)[/tex] at this point.

To find the minimum value of the quadratic function [tex]\( f(x) = x^2 + 5x + 6[/tex], we can use the vertex formula. The vertex of a quadratic function [tex]\( ax^2 + bx + c \) is given by the point \( \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right) \).[/tex]

[tex]For \( f(x) = x^2 + 5x + 6 \), \( a = 1 \), \( b = 5 \), and \( c = 6 \).[/tex]

The x-coordinate of the vertex is [tex]\( -\frac{b}{2a} = -\frac{5}{2(1)} = -\frac{5}{2} \).[/tex]

To find the minimum value, we substitute [tex]\( x = -\frac{5}{2} \)[/tex] into the function:

[tex]\[ f\left(-\frac{5}{2}\right) = \left(-\frac{5}{2}\right)^2 + 5\left(-\frac{5}{2}\right) + 6 \]\[ = \frac{25}{4} - \frac{25}{2} + 6 \]\[ = \frac{25}{4} - \frac{50}{4} + \frac{24}{4} \]\[ = \frac{-1}{4} \][/tex]

So, the minimum value of the function occurs at [tex]\( x = -\frac{5}{2}[/tex], and the function reaches its minimum value of [tex]\( -\frac{1}{4} \)[/tex] at this point.