Which of the following are miscellaneous taxes? Gasoline tax License plate fees Hunting license fees Inheritance tax Marriage license fees

General Idea:

Solving for a variable means getting that variable by itself by UNDOING whatever is done to it. We need to perform reverse operation to UNDO.

Applying the concept:

[tex] \frac{4q-6n}{3m}=f\\ \\ To \; undo \; 3m\; in\; the \; denominator \; 3m, \; MULTIPLY \; 3m\; on\; both\; sides [/tex]

[tex] 3m \times \frac{4q-6n}{3m}=f \times 3m\\ \\ 4q-6n=3mf\\Add \; 6n \; both\; sides\\ \\ 4q-6n+6n=3mf+6n\\ Combine \; like\; terms\\ \\ 4q=3mf+6n\\Divide \; 4\; on\; both\; sides\\ \\ \frac{4q}{4} =\frac{3mf+6n}{4} \\ Simplify\; fraction\; in\; left\; sides\\ \\ q=\frac{3mf+6n}{4} [/tex]

Conclusion:

[tex] Solving \; the \; equation \; \frac{4q-6n}{3m}=f \; for q, \; we get...\\ \\q=\frac{3mf+6n}{4} [/tex]

Answer:

a 24

Step-by-step explanation:

By applying Fermat's Little Theorem, we can find the least positive residue of [tex]2^{1000000[/tex] modulo 17. Using the repeated squaring method, we calculate that [tex]2^{16[/tex] is congruent to 1 modulo 17. Therefore, [tex]2^{1000000[/tex] is also congruent to 1 modulo 17.

Fermat's Little Theorem states that if p is a prime number and a is any positive integer that is not divisible by p, then a raised to the power of p-1 is congruent to 1 modulo p. In this case, we need to find the least positive residue of 2 raised to the power of 1000000 modulo 17.

First, we need to find the value of 2 raised to the power of 16 modulo 17, since 17 is a prime number. Using the repeated squaring method, we can calculate:

[tex]2^2[/tex] = 4 (mod 17)

[tex]2^4[/tex] = [tex](2^2)^2[/tex] = [tex]4^2[/tex] = 16 (mod 17)

[tex]2^8[/tex] = [tex](2^4)^2[/tex] = [tex]16^2[/tex] = 1 (mod 17)

[tex]2^{16[/tex] = [tex](2^8)^2[/tex] = [tex]1^2[/tex] = 1 (mod 17)

Now, we can find the value of [tex]2^{1000000[/tex] modulo 17 using the fact that [tex]2^{16[/tex] is congruent to 1 modulo 17:

[tex]2^{1000000[/tex] = [tex](2^{16})^{62500[/tex] x [tex]2^0[/tex] = [tex]1^{62500[/tex] x 1 = 1 (mod 17)

Therefore, the least positive residue of [tex]2^{1000000[/tex] modulo 17 is 1.

Answer:

Option C is correct

(1, 5)

Step-by-step explanation:

A system of equations contains two or more than two linear equations that divide two or more unknowns.

To find a solution for the system of equations,

we must find a value from the graph that is true for all equations in the given system.

We know that:

If the graphs of the equations intersect,

then there is one solution that is true for both equations.

Given the graph.

There are two lines

Equation of Blue line : y = -x+6

Equation of Red line = y = 2x+3

Since, the graph of two lines y = -x+6 and  y = 2x+3 intersects which means there is one solution that satisfies the both equation.

Therefore, the solution to the system of equations graphed is, (1, 5)

Final answer:

Using the z-score calculation and the standard normal distribution, approximately 6.68% of fully grown English oak trees are taller than 110 feet.

Explanation:

To find the proportion of fully grown English oak trees taller than 110 feet, given that the heights are normally distributed with a mean of 95 feet and a standard deviation of 10 feet, we first calculate the z-score for a height of 110 feet. The z-score is calculated using the formula: z = (X - μ) / σ, where X is the value of interest (110 feet), μ is the mean (95 feet), and σ is the standard deviation (10 feet).

By substituting the given values: z = (110 - 95) / 10 = 1.5. This z-score represents the number of standard deviations 110 feet is above the mean. To find the proportion of trees taller than 110 feet, we consult the standard normal distribution table or use a calculator that provides the area to the right of the z-score, which corresponds to the proportion we seek.

The standard normal distribution table or a calculator shows that the proportion of the area under the curve to the right of a z-score of 1.5 is approximately 0.0668.

Thus, about 6.68% of fully grown English oak trees are taller than 110 feet.

Polygons can be classified by their sides and angles using formulas and definitions like regular polygons with equal sides and angles.

Polygons can be classified by their sides and angles. A regular polygon has equal sides and equal angles, such as an equilateral triangle or a square. The sum of interior angles in any polygon can be calculated using the formula 180 * (n - 2) degrees where n is the number of sides.

Answer:

Step-by-step explanation:

Answer:

  690

Step-by-step explanation:

Put 30 where m is in the expression and do the arithmetic. Or, simpify the expression first, then do the same.

  14·30 +30 +17·30 -9·30 = 420 +30 +510 -270 = 690

Simplifying first, you have ...

  14m +m +17m -9m = m(14 +1 +17 -9) = 23m

  23·30 = 690

To solve the equation 14m + m + 17m - 9m when m=30, substitute the value of m in the equation and follow order or operations. The result is 960.

To solve for the given problem involving the variable m, simply substitute the value of m into the equation. The equation given is 14m + m + 17m - 9m. Substituting 30 for m in the equation gives us: 14(30) + 30 + 17(30) - 9(30). To solve this, use the order of operations principle, which states that multiplication and division should be performed before addition and subtraction.

First, perform the multiplication: 420 + 30 + 510 - 270. Next, add and subtract from left to right: 960. Thus, if m = 30, then 14m + m + 17m - 9m = 960.

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HL congruence theorems or postulates used to prove ΔLMN ≈ΔOPQ.

Congruent figures have sides that are the same length and angles that are the same measurement. In other words, congruent figures are those in which one figure superimposes another.

If two line segments have the same length, they are said to be congruent. If two angles have the same measure, they are said to be congruent. If the matching sides and angles of two triangles are equal, they are said to be congruent.

We have two Triangles as ΔLMN and ΔOPQ

As, per the figure both triangles are Right angles Triangle.

Also, LM = OP

MN = PQ

LN = OQ

and, the hypotenuse of both triangles also Equal.

Thus, both triangle are Congruent using HL theorem.

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The orbital period (in years) of an asteroid whose average distance from the sun is 10 AU is 212314723 years.

Orbital period of a body is the the time taken by a body to cover one circular path around the central object under whose influence it is following circular path.

Given is a asteroid moving around the sun.

Mean distance from the Sun [x] = 10 AU = 1.496 x 10¹² m.

Assume that the time period of the asteroid is T.

The formula to calculate the orbital period around the sun is -

T²/ R³ = 4π²/GM[s]

Where -

R is the mean distance from sun.

G is the universal Gravitation constant.

M[s] is the mass of Sun.

Substituting the values, we get -

T² = (1.496 x 10¹² x 4 x 3.14 x 3.14)/(6.673 x 10⁻¹¹ x 1.98 x 10³⁰)

T² = (59 x 10¹²)/(13.2 x 10⁻¹⁹)

T² = 4.45 x 10 x 10³⁰

T² = 44.5 x 10³⁰

T = √44.5 x √10³⁰

T = 6.7 x 10¹⁵ seconds

In years, the time period will be -

T = 212314723 years (approx.)

Therefore, the orbital period (in years) of an asteroid whose average distance from the sun is 10 AU is 212314723 years.

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The modulo operator determines the remainder of a division operation. For the given expressions, the results are a. 1, b. 4, c. 2, d. 6

The question is asking to evaluate several expressions using the modulo operation. Modulo is a mathematical operation that finds the remainder or signed remainder of a division, after one number is divided by another (called the modulus). Here's how we can solve each:

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Answer:

8.2%

Step-by-step explanation:

o-ware

Using the binomial distribution, it is found that there is a 0.936 = 93.6% probability that he makes 2 shots or less.

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For each throw, there are only two possible outcomes. Either the player makes it, or he misses it. The probability of making a throw is independent of any other throw, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this question:

The probability of making 2 or less is the probability that he does not make all of them, that is:

[tex]P(X < 3) = 1 - P(X = 3)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{3,3}.(0.4)^{3}.(0.6)^{0} = 0.064[/tex]

[tex]P(X < 3) = 1 - P(X = 3) = 1 - 0.064 = 0.936[/tex]

0.936 = 93.6% probability that he makes 2 shots or less.

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