•15 POINTS• don’t need to explain.

a.)
b.)
c.)

The probability of the mean height of the sample of boys being at least 6 cm higher than the mean height of the sample of girls can be calculated using a two-sample t-test. The t-test compares the means of two samples and determines whether they are different from one another. The t-test requires the assumption of equal variances and normal distribution of the data.

What does the math term "probability" mean?

Simply put, probability is the likelihood that something will occur. When we don't know how an event will turn out, we can discuss the likelihood or likelihood of several outcomes. Statistics is the study of events that follow a probability distribution.

Given that the variances are equal, the t-test statistic can be calculated as:

t = (mean_boys - mean_girls - 6) / (sqrt(variance/n))

where n is the sample size of each group (8 in this case).

The probability of observing a t-value as extreme or more extreme than this value can be calculated using the cumulative distribution function (CDF) of a t-distribution with n1 + n2 - 2 degrees of freedom.

Because you are looking for the probability that the mean height of the sample of boys is at least 6cm higher than the mean height of the sample of girls, this is a one-tailed test. The probability of a one-tailed test can be calculated by subtracting the probability of a two-tailed test from 1.

The probability can be calculated using a t-distribution table with 14 d.f. or using software such as R or excel.

It is important to note that the sample size is very small, with 8 boys and 8 girls, which may not be representative of the population. As sample size increases the probability of getting a significant difference increases.

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Answer: 85 seconds

Step-by-step explanation: 12.75/0.15=85

The equation is

log ( milligrams of drug A ) =3 - 0.0737 ( Time)

As per the details  stated above,

Curved relationship between the number of milligrams of drug A is  (x) still in the body and the number of hours since drug A was delivered (y)

The co-efficient of the constant is 3.00

the co-efficient of the time is -0.0737

the SE  co-efficient of the constant is 0.4017

the SE co-efficient of the time is 0.005

the t-ratio of the constant is 7.481

the t-ratio of the time is -14.74

now the equation formed is

logarithm of drug A in milligrams is  (The co-efficient of the constant - the co-efficient of the time )

that is log ( milligrams of drug A ) =3 - 0.0737 ( Time).

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The Filling time is After 25 minutes, both Aquariums will contain same amount of water.

you've fully prepared the aquarium, all you need to do is follow these quick and easy steps and safely fill your aquarium with water.

Calculate the Filling Time of the aquarium Explanation

Given,

Water in Aquarium A = 4.6 gallons

Water in Aquarium B = 54.6 gallons

Filling rate = 1.2 gallons per minute.

Draining rate = 0.8 gallons per minute.

Let,

x represents the minutes.

A(x) = 4.6 + 1.2x

(Because Luis is adding water in the tank.)

B(x) = 54.6 - 0.8x

(Isacc is draining water from the tank.)

For same amount of water;

A(x) = B(x)

4.6+1.2x=54.6-0.8x

1.2x+0.8x=54.6-4.6

2x=50

Dividing both sides by 2

2x/2=50/2

x=25

Therefore, After 25 minutes, both Aquariums will contain same amount of water.

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The approximate length, in feet, of lights that she needs to create one triangle is 64

This is an approximate estimate as the exact length would depend on the specific measurements of the roof line. However, if the roof line is a typical house roof and the string lights are hung horizontally across to connect the ends of the roof line, creating a triangle, the approximate length of lights needed would be the hypotenuse of the triangle (the longest side) which is equal to the square root of (leg 1^2 + leg 2^2).

A typical house roof is approximately 40 feet wide and 20 feet tall, which creates a right triangle with legs of 40 and 20. The hypotenuse of this triangle (the length of lights needed) would be approximately 64 feet (the square root of (40^2 + 20^2) = 64).

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The equation of the perpendicular bisector of the segment with the endpoints (-8,-1) and (4,5)  is 2x+y = -2.

The line between these locations' slopes must equal the negative reciprocal of the bisector's slope. It must go through the segment midway.

The slope of the line through the given points :

m = (y2 -y1)/(x2 -x1) = (5 -(-1))/(4 -(-8)) = 6/12 = 1/2

Then, The slope of the required bisector:   m = -1/(1/2) = -2

The midpoint of the given segment:

((-8, -1) +(4, 5))/2 = (-8+4, -1+5)/2 = (-4, 4)/2 = (-2, 2)

Then the point-slope form of the equation of the bisector:

y -y1 = m(x -x1)

y -2 = -2(x -(-2))

y = -2x -4 +2

y = -2x -2 . . . . . . . slope-intercept form equation

2x +y = -2 . . . . . . .standard form equation

Thus, The equation of the perpendicular bisector of the segment with the endpoints (-8,-1) and (4,5)  is 2x+y = -2.

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The steps to solve two polynomial equations are given below.

What is polynomial?

A polynomial is a mathematical expression made up of coefficients and indeterminates that uses only the operations addition, subtraction, multiplication, and powers of positive integers of the variables.

x^2 + 4x + 7 is an illustration of a polynomial with a single indeterminate x.

Apply the Zero Factor Property to an Equation Solving.

ZERO. FACTOR, write the equation with one side equal to zero. the expression by a factor.

PROPERTY. Solve the equation by setting each factor to zero.

Check by adding substitutes to the initial equation.

Write a polynomial equation in standard form before attempting to solve it. Factor it, then set each variable factor to zero once it has reached zero. The original equations' solutions are the solutions to the resulting equations. Factoring cannot always be used to solve polynomial equations.

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Answer:

3.23, -1.23

Step-by-step explanation:

Draw the line y=1

See where it intersects x^2-2x-3

The x coordinates of the intersections are your answer

In scenerio 1, the money in the retirement account at 62 years old is         $1,267,250.76

In scenerio 2, the money in the retirement account at 62 years old is         $ 633,625.38.

Scenerio 1:

a) 6% of salary = Salary * 6% = $6,100 * 6% = $366

b) Total amount the company paid monthly = 6% of salary = Salary * 6% = $6,100 * 6% = $366

c) The money in the retirement account at 62 years old can be calculated using the formula for calculating the Future Value (FV) of an Ordinary Annuity given as follows:

FV = M * (((1 + r)^n - 1) / r) ................................. (1)

Where:

FV = Future value = Money in the retirement account at 62 years old = ?

M = Monthly payment = Contribution you + Contribution by the employer = $366 + $366 = $732

r = Monthly interest rate = Annual interest rate / 12 = 6.28% / 12 = 0.0628 / 12 = 0.0052

n = number of months = Number years * Number of months in a year = (Retirement age – Age when hired) * 12 = (62 - 25) * 12 = 37 * 12 = 444

Substituting all the values into equation (1), we have:

FV = $732 * (((1 + 0.0052)^444 - 1) / 0.0052)

FV = Money in the retirement account at 62 years old = $1,267,250.76

d) Amount of money you put into the account = 6% of salary * number of months = $366 * 444 = $162,504

e) Amount of money your employer put into your account = Amount of money you put into the account = $162,504

f) Amount your account gained in interest over the years = FV – (Amount of money you put into the account + Amount of money your employer put into your account) = $1,267,250.76 – ($162,504 + $162,504) = $942,242.76

Scenario 2:

a) 3% of salary = Salary * 3% = $6,100 * 3% = $183

b) Total amount the company paid monthly = 3% of salary = Salary * 3% = $6,100 * 3% = $183

c) Using equation (1), we have:

M = Monthly payment = Contribution you + Contribution by the employer = $183 + $183 = $366

Substituting all the values into equation (1), we have:

FV = $366 * (((1 + 0.0052)^444 - 1) / 0.0052)

FV = Money in the retirement account at 62 years old = $633,625.38

d) Amount of money you put into the account = 3% of salary * number of months = $183 * 444 = $81,252.00

e) Amount of money your employer put into your account = Amount of money you put into the account = $81,252

f) Amount your account gained in interest over the years = FV – (Amount of money you put into the account + Amount of money your employer put into your account) = $633,625.38 – ($81,252 + $81,252) = $471,121.38

The difference in the amount of money at age 62 between scenarios 1 and 2 = Money in the retirement account at 62 years old in scenarios 1 + Money in the retirement account at 62 years old in scenarios 2 = $1,267,250.76 - $633,625.38 = $633,625.38

Difference in the amounts of money you put into retirement between scenarios 1 and 2 = $366 - $183 = $183

Amount got from your employer in scenario 1 than 2 = $366 - $183 = $183

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The graph of the linear equation 2x + 3y = 6 is a straight line that cuts the Y-axis at the point (0,2).

A linear equation is an equation in which the highest power of the variable is 1.

The general form of a linear equation is:ax + by = c where a, b and c are constants and x and y are variables. This equation represents a straight line when plotted on a graph.

The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line and b is the y-intercept

To find the points at which the line will intersect y-axis,

take x=0 and substitute in the linear equation 2x+3y=6

2(0) + 3y = 6

3y = 6

y = 2

So the point where the line cuts the Y-axis is (0,2)

The straight line of the linear equation 2x+3y=6 intersects the y-axis at (0,2).

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To prove that ΔABC is isosceles.

An isosceles triangle is a type of triangle in which the length of its two sides, and measure of its two internal angles are equal. When a bisector of an angle is drawn to one of the internal angles of the triangle, two congruent triangles of corresponding properties are produced.

To prove that ΔABC is isosceles, we have:

<CAD ≅ <BAD                             (definition of an angle bisector)

CD ≅ BD                                      (definition of a midpoint)

AC ≅ AB                                      (congruent sides of an isosceles)

<ACD ≅ <ABD                             (congruent internal angles of an isosceles)

<ADC ≅ <ADB                             (definition of right angle)

Therefore it can be concluded that;

ΔABC is an isosceles triangle    (Angle-Side-Side theorem)

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Answer:

Step-by-step explanation:

You want the cost of each item when person A buys 10 granola bars and 6 cups of yogurt for $18, and person B buys 5 granola bars and 4 cups of yogurt for $9.50.

The equations describing the relations can be written ...

  10g +6y = 18 . . . . . . . person A's purchase

  5g +4y = 9.50 . . . . . . person B's purchase

Subtracting half of the first equation from the second, we find ...

  (5g +4y) -1/2(10g +6y) = (9.50) -1/2(18)

  y = 0.50 . . . . . . . . . . simplify

Substituting this into the first equation gives ...

  10g +6(0.50) = 18

  10g = 15 . . . . . . . . . . subtract 3

  g = 1.50 . . . . . . . . . divide by 10

Each granola bar costs $1.50; each cup of yogurt costs $0.50.

__

Additional comment

The attachment shows the solution found using a calculator to find the reduced row-echelon form of the augmented coefficient matrix. The variable values are in the right column of the reduced matrix. This tells us granola bars are $1.50, and yogurt cups are $0.50, as we found above.

Step-by-step explanation:

12 holes / 6 minutes = 2 holes / 1 minute

the simplified ratio is 2/1 = 2.

so, for 25 minutes we need to keep the same ratio :

x holes / 25 minutes = x/25 = 2

x = 2×25 = 50

it drills 50 holes in 25 minutes.

Answer:

(I cannot show a picture of a bar model, apologies)

∠1 and ∠3 are vertical angles and congruent.

What is congruence?

In geometry, two figures or objects are said to be congruent if their shapes and sizes match, or if one is the mirror image of the other.

What are vertical angles?

The opposing angles form when two lines cross. They are constantly on par. In this illustration, the angles a° and b° are vertical. The term "vertical" refers to the vertex, not up or down, where they cross.

Here,

we have given a diagram and we have to determine which statement is true that satisfies the true conditions.

By applying congruent property. we get

∠1 = ∠3 are vertivally opposite angle.

This is the only condition that satisfy the true condition from given statement.

Hence, ∠1 and ∠3 are vertical angles and congruent.

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Answer:

93.3% i hope this is right

Step-by-step explanation:

percentage increase= final-original/originalx100

58-30/30x100=93.3333%

Answer:

[tex]C=59.6902604182[/tex]

Step-by-step explanation:

The formula to calculate the circumference of a circle given the diameter is:

[tex]C=\pi d[/tex]

Substitute and solve:

[tex]C=\pi \times19=59.6902604182[/tex]
[tex]C=59.6902604182[/tex]

Since the question doesn't say to round your answer is "59.6902604182."

Hope this helps.

At this rate, the ratio or proportion of red balloons to the total balloons, the number of red balloons out of 48 is 6.

Proportion describes the two or more ratios equated to each other.

Proportions are depicted as fractional values in decimals, fractions, or percentages.

In this situation, we equate the ratio or proportion of red balloons to the sample to determine the number that will likely be red.

The total number of balloons thrown in a water balloon fight = 6,960

The number of the balloons that are red = 870

The proportion of the red balloons to the total = 0.125 or 12.5% (870/6,960)

In a sample of 48 balloons, the number of red balloons will be 6 (48 x 0.125)

Thus, out of 48 balloons, there are 6 red balloons at this constant rate.

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The basis for the eigenspace corresponding to lambda=5,1,4 are None,[tex]\left[\begin{array}{c}-1 \\\frac{1}{2} \\0\end{array}\right][/tex] and [tex]$\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]$[/tex]

[tex]$$A=\left[\begin{array}{ccc}5 & -12 & 10 \\0 & 7 & -3 \\0 & 6 & -2\end{array}\right]$$[/tex]

Eigenspace corresponding to lambda=5,1,4

The eigenspace E_lambda corresponding to the eigenvalue lambda is the null space of the matrix a [tex]\mathrm{A}-(\lambda) \mathrm{I}"[/tex]

for lambda=5

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-5 \mathrm{I})$$[/tex]

Reducing the matrix A-5I by elementary row operations

[tex]$$\begin{aligned}A-5 I & =\left[\begin{array}{ccc}5-5 & -12 & 10 \\0 & 7-5 & -3 \\0 & 6 & -2-5\end{array}\right] \\& =\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 2 & -3 \\0 & 6 & -7\end{array}\right] \\& \sim\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_2 \rightarrow \frac{R_2}{2} \\& \sim\left[\begin{array}{ccc}1 & 0 & -8 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_1 \rightarrow R_1+2 R_2\end{aligned}$$[/tex]

[tex]\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 2\end{array}\right] R_3 \rightarrow R_3-6 R_2$$\\\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] R_3 \rightarrow \frac{\mathrm{R}_3}{2}$$\\\sim\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+8 \mathrm{R}_3$[/tex]

[tex]$\sim\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] R_2 \rightarrow R_2+\frac{2 R_3}{2}$[/tex]

The solutions x of A-5I=0 satisfy x_1=x_2=x_3=0 that is, the null space solves the matrix

[tex]$$\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

Hence The null space is [tex]\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] E_5[/tex] has no basis

[tex]$$\begin{aligned}& \text { case: } 2 \\& \text { for } \lambda=1 \\& \mathrm{E}_5=\mathrm{N}(\mathrm{A}-(1) \mathrm{I})\end{aligned}$$[/tex]

we reduce the matrix A-I by elementary row operations as follows.

[tex]$$\begin{aligned}A-1 & =\left[\begin{array}{ccc}5-1 & -12 & 10 \\0 & 7-1 & -3 \\0 & 6 & -2-1\end{array}\right] \\& =\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 6 & -3 \\0 & 6 & -3\end{array}\right] R_1 \rightarrow \frac{R_1}{4} \\& \sim\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 1 & -\frac{1}{2} \\0 & 6 & -3\end{array}\right] R_2 \rightarrow \frac{R_2}{6}\end{aligned}[/tex]

[tex]$$$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 6 & -3\end{array}\right] R_1 \rightarrow R_1+3 R_2$\\$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right] R_3 \rightarrow R_3-6 R_2$[/tex]

Thus, the solutions x of (A-I) X=0 satisfy

[tex]$\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=-\mathrm{t}, \mathrm{x}_2=\frac{\mathrm{t}}{2}$[/tex]

[tex]$\vec{x}=\left[\begin{array}{c}-t \\ \frac{t}{2} \\ t\end{array}\right]=\left[\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right] t$[/tex]

The Basis for the nullspace A-I will be: [tex]$\left.\left(\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right]\right)$[/tex]

case:3

lambda=4

[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-(4) \mathrm{I})$$[/tex]

we reduce the matrix A-4I by elementary row operations as follows.

[tex]$\begin{aligned} A-4 \mid & =\left[\begin{array}{ccc}5-4 & -12 & 10 \\ 0 & 7-4 & -3 \\ 0 & 6 & -2-4\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 3 & -3 \\ 0 & 6 & -6\end{array}\right] \\ & \sim\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] R_2 \rightarrow \frac{R_2}{3}\end{aligned}$[/tex]

[tex]$\begin{aligned} & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+12 \mathrm{R}_2 \\ & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-6 \mathrm{R}_2\end{aligned}$[/tex]

Thus, the solutions x of (A-4IX)=0 satisfy

[tex]$$\left[\begin{array}{ccc}1 & 0 & -2 \\0 & 1 & -1 \\0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]

x_3=t

[tex]$\Rightarrow \mathrm{x}_1=2 \mathrm{t}, \mathrm{x}_2=\mathrm{t}$[/tex]

[tex]$$\vec{x}=\left[\begin{array}{c}2 t \\t \\t\end{array}\right]=\left[\begin{array}{l}2 \\1 \\1\end{array}\right] t$$[/tex]

The Basis for the nullspace A-4 I will be [tex]\left(\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]\right)[/tex]

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Latitude 36°30′ was designated by the Missouri Compromise of 1820 as the northern limit for the legalization of slavery in the western territories.

By Stephen A. Douglas's Kansas-Nebraska Act of 1854, the Missouri Compromise's prohibition on slavery in the former Louisiana Territory north of the parallel 36°30′ north had been essentially removed. Latitude 36°30′ was designated by the Missouri Compromise of 1820 as the northern limit for the legalization of slavery in the western territories.

The expansion of slavery into the western lands was addressed by the Missouri Compromise, the Compromise of 1850, and "Bleeding Kansas." The balance between "free" and "slave" states in the Union was a factor in each of them.

Slavery was proposed to be outlawed in the remaining Louisiana Territory above the 36° 30' latitude line in the Missouri Compromise. The Kansas-Nebraska Act of 1854 abolished this clause after it had been in effect for 34 years.

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The Missouri Compromise of 1820 established latitude 36°30′ as the northern boundary for the legalisation of slavery in the western territories.

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The bake sale sign has a 6 ft2 space.

An object's area is how much space it takes up in two dimensions.

It is the measurement of the quantity of unit squares that completely cover the surface of a closed figure.

The word "area" refers to a free space.

A shape's length and width are used to compute its area. Unidimensional length is expressed in terms of feet (ft), yards (yd), inches (in), etc.

But a shape's area is a two-dimensional measurement. In order to measure something in a square, it is done so using measurements like square inches (in2), square feet (ft 2), square yards (yd2), etc.

The majority of forms and things have corners and edges. 

According to our question-

The opposite sides of a rectangle's quadrilateral (it has four sides and four angles) are equal and parallel.

Area of a rectangle equals length times width.

The area of the bake sale sign = length * breadth

Hence, The bake sale sign has a 6 ft2 space.

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Each friend of Arya will receive 0.625 cupcakes.

To divide 5 cupcakes equally among 8 friends, we can use simple division. We would divide the total number of cupcakes (5) by the total number of friends (8) to find the number of cupcakes each friend would receive. In this case, 5 divided by 8 is equal to 0.625. This means that each friend will receive 0.625 cupcakes. However, since cupcakes are not divisible it is not possible to divide 5 cupcakes equally among 8 friends. In this case, either Arya will have to make more cupcakes or they will have to decide on a different way to divide the cupcakes.

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Answer:

y = -3x +20

Step-by-step explanation:

Equation of line: y =mx + b

Where m is the slope and b is the y-intercept.

3x + y = 17

Write the above equation in y = mx +b form.

      y = -3x + 17

m = -3

Parallel lines have same slope.

So, slope of the required line is (-3).

     y = -3x + b

This line is passing through (4,8). Substitute in the above equation and find the value of 'b'.

       8 = -3*4 + b

        8 = -12 + b

  8 + 12 = b

          b = 20

Equation of the line:

               [tex]\sf \boxed{ y = -3x + 20}[/tex]

The displacement of the pizza delivery driver for the full journey is therefore 6.08 miles.

The pizza delivery driver's displacement for the entire trip is the total distance and direction that she travels from the restaurant to her last delivery and back.

To find her displacement, we can use the Pythagorean theorem to find the total distance from the start to the endpoint.

The displacement is the distance between the start and end point and the direction of the trip. Here we can see that she traveled 4 miles north, 6 miles east, and 5 miles south,

To find the displacement we will use the Pythagorean theorem (ax2 + bx2 = cx2)

where c is the displacement and a and b are the east-west and north-south distances respectively.

so, a = 6 miles (east-west)

b = (4 miles - 5 miles) = -1 miles (north-south)

so, c = √(ax2 + bx2)

c = √(6x2 + (-1)x2)

c = √(36 + 1)

c = √(37)

c = 6.08 miles

So the pizza delivery driver's displacement for the entire trip is 6.08 miles.

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The height of the equilateral triangle when its side length is given as 8 inches is calculated to be 6.93 inches.

The equilateral triangle has all sides and angles equal. The angles are always equal to 60 degrees.

Given that,

Side length of the equilateral triangle = 8 inches

Angles = 60 degrees

Height = ?

Therefore, let us find the height of the triangle using the Pythagoras theorem,

c² = a² + b²

8² - 4² = b²

b² = 64 - 16

b² = 48

b = 6.93 inches

Thus, the height of the equilateral triangle is calculated to be 6.93 inches.

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The value of DX for solving simultaneous equations 3x 2y =- 11 and 7x 4y 9 by Cramer's rule is -13/5.

The general formula for Cramer's rule is

DX = detX/detA

Where detX is the determinant of the matrix formed by replacing the x-column of the original matrix with the constant column vector, and detA is the determinant of the original matrix.

For the given equations,

A = [3, 2; 7, 4]

X = [-11; -9]

Therefore,

detA = 3*4 - 2*7 = -10

detX = 3*(-9) - 2*(-11) = 13

Hence, DX = detX/detA = 13/-10 = -13/5

The value of DX for solving simultaneous  

3x 2y =- 11 and 7x 4y 9 by Cramer's rule equation is -13/5.

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to get the slope of any straight line, we simply need two points off of it, let's use those two in the picture below.

[tex](\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{3}-\stackrel{y1}{(-4)}}}{\underset{\textit{\large run}} {\underset{x_2}{4}-\underset{x_1}{(-2)}}} \implies \cfrac{3 +4}{4 +2} \implies {\large \begin{array}{llll} \cfrac{7 }{ 6 } \end{array}}[/tex]