What is the greatest common factor of 12 and 42

Answer:

A black-tailed jackrabbit can hopes 36 in 5 seconds.

Step-by-step explanation:

Consider the provided information.

Black tailed jackrabbit hopes 51 feet in one second.

Therefore, in 5 second Black tailed jackrabbit covers a distance of:

51×5=255  feet

So, in 5 seconds a black-tailed jackrabbit will travel 255 feet.

In one hops Black tailed jackrabbit covers 7 feet.

Now find the number of hops, we need to divide the distance by the length of one hop:

[tex]255\div 7=\frac{255}{7}\approx36[/tex]

Hence, A black-tailed jackrabbit can hopes 36 in 5 seconds.

Answer:

Rational.

Step-by-step explanation:

We are asked to determine the type of number 0.5.

We can see that our given number is a decimal.

We can also see that our given number is a terminating decimal, so we can represent it as a fraction.

[tex]0.5\times \frac{10}{10}=\frac{5}{10}=\frac{1}{2}[/tex]

Therefor, our given number is a rational number as we can represent it as a fraction.

We have that if the sin 30 degrees is 1/2 then the cos __60=1/2.__

Option C

Question Parameters:

The sin 30 degrees is 1/2

Generally, the Trigonometry of a right angle triangle gives values for sin, cos, tan.

Using the triangle we have that

if the sin 30 degrees is 1/2 then the cos60=1/2

Therefore, if the sin 30 degrees is 1/2 then the cos __60=1/2__.

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The two numbers are 8 and 12.

What is Algebraic expression ?

Algebraic expressions are numbers expressed using letters or alphabets without specifying their values. Algebra taught us how to express unknown values using letters such as x, y, and z. There can also be a single value that is placed before and multiplied by a variable in an algebraic expression in addition to these letters, which are called variables.

Let us assume first number be = x

and, let the second number be = y

the,

One number is 4 less than a second number means :

x = y - 4........(1)

Twice the second number is 16 less than 5 times the first means :

2y = 5x - 16.......(2)

Substitute x equals from the first equation in the second equation.

2y = 5(y - 4) - 16

2y = 5y - 20 - 16

-3y = -36

y = 12

Now use the first equation for find x.

x = y - 4

x = 12 - 4

x = 8

Thus, The numbers are 8 and 12.

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Final answer:

To find the value of k so that the line containing the points (-5,k) and (2,10) is parallel to the line containing the points (5,5) and (1,-4), we need to find the slope of the second line and use it to find the value of k.

Explanation:

To find the value of k so that the line containing the points (-5, k) and (2, 10) is parallel to the line containing the points (5, 5) and (1, -4), we need to find the slope of the second line and use it to find the value of k. Firstly, let's find the slope of the second line:

The formula we apply is the slope

= (y2 - y1) / (x2 - x1).

slope = -9 / -4 = 9/4;

slope = (-4 - 5) / (1 - 5)

The lines now have the same slope since they are parallel.

So, the slope of the line passing through (-5, k) and (2, 10) should also be 9/4. We can find the slope using the formula:

slope = (y2 - y1) / (x2 - x1)

Substituting the coordinates (-5, k) and (2, 10):

9/4 = (10 - k) / (2 - (-5))

By simplifying the equation, we get:

9 = 7(10 - k)

Now, let's solve for k:

90 - 9k = 63

-9k = 63 - 90

-9k = -27

k = 3

To sketch a triangle with an acute angle θ and cot(θ) = 2, we can choose an angle in the first quadrant, such as 30 degrees, and draw a right triangle with sides 2 and 1. The hypotenuse can be found using the Pythagorean theorem.

To sketch a triangle with an acute angle θ and cot(θ) = 2, we can use the fact that cotangent is the reciprocal of the tangent function, so tan(θ) = 1/2. Since tangents are positive in the first and third quadrants, we can consider an angle in the first quadrant. Let's choose the angle as 30 degrees.

Now, we can draw a right triangle with one of the acute angles as 30 degrees. Since tan(θ) = opposite/adjacent, we can choose the side opposite the angle as 1 and the side adjacent as 2. Therefore, the triangle has sides 2, 1, and the hypotenuse can be found using the Pythagorean theorem.

The triangle has an acute angle θ and cot(θ) = 2. It has sides of length 2, and 1, and the hypotenuse can be found using the Pythagorean theorem.

The slope of a line perpendicular to the given line is -5/2.

The equation of the line given is 5y = 10 + 2x. To find the slope of this line, we need to rearrange the equation into the slope-intercept form, y = mx + b, where m represents the slope. Dividing both sides of the equation by 5, we get y = 2/5x + 2. Comparing this equation to the slope-intercept form, we can see that the slope is 2/5.

To find the slope of a line perpendicular to this line, we use the fact that the product of the slopes of two perpendicular lines is -1. Therefore, the slope of the perpendicular line is the negative reciprocal of the slope of the given line. The negative reciprocal of 2/5 is -5/2.

Therefore, the slope of a line that is perpendicular to the line with the equation 5y = 10 + 2x is -5/2.

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The expected recorded score when the total number of points earned on the two parts is recorded is 20.4.

The expected recorded score when the maximum of the two scores is recorded is 25.7.

We have,

(a)

If the score recorded in the grade book is the total number of points earned on the two parts (x + y):

We will calculate the expected value (E) using the joint probability mass function (pmf) values given in the table.

E(x + y) = Σ[(x + y) * p(x, y)]

Using the table values:

E(x + y) = (0 * 0.02) + (5 * 0.06) + (10 * 0.02) + (15 * 0.10) + (20 * 0.04) + (25 * 0.13) + (30 * 0.20) + (35 * 0.10) + (40 * 0.01) + (45 * 0.15) + (50 * 0.16) + (55 * 0.01)

E(x + y) = 20.4

(b)

If the maximum of the two scores is recorded:

We need to find the maximum value between x and y for each combination of scores in the table and calculate the expected value using the joint pmf.

E(max(x, y)) = Σ[max(x, y) * p(x, y)]

Using the table values:

E(max(x, y)) = (0 * 0.02) + (5 * 0.06) + (10 * 0.15) + (15 * 0.20) + (20 * 0.20) + (25 * 0.15) + (30 * 0.15) + (35 * 0.15) + (40 * 0.16) + (45 * 0.16) + (50 * 0.16) + (55 * 0.01)

E(max(x, y)) = 25.7

Therefore,

The expected recorded score when the total number of points earned on the two parts is recorded is 20.4.

The expected recorded score when the maximum of the two scores is recorded is 25.7.

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The expected recorded score in the grade book (E(X+Y)) is calculated by multiplying each possible total score by its probability and summing these products. The expected maximum score is calculated similarly, but considers the probabilities of each possible maximum score. Application of these principles requires access to full probability mass function data.

The student's question relates to the calculation of expected values in the context of a two-part quiz, with scores registered as the total (x+y) or the maximum of the two scores. Given a joint probability mass function (pmf) for x and y, we can calculate expected scores in both cases.

Generally, the expected value E(X) of a discrete random variable X is calculated by multiplying each possible value of X by its probability, and then summing these products. This is represented by the formula E(X) = Σ xP(x). Assuming the same logic for the expected total score on the quiz, E(X+Y) would be calculated by summing the products of each possible total score and its probability.

To find the expected maximum score, we'd perform a similar calculation, but consider the probabilities of each possible maximum score instead.

Without explicit probabilities in the question, no exact solutions can be provided. However, the processes outlined here would allow the student to solve such problems with access to the full pmf data.

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62 +62 = 124

360 - 124 = 176

176/2 = 88

BED = 88 degrees