Which one of the following statements about quadrilaterals is true?

A.) A quadrilateral with only one set of parallel sides is a square.
B.) A quadrilateral with a right angle is a rhombus.
C.) A quadrilateral with two pairs of parallel sides is a trapezoid.
D.) A parallelogram with a right angle is a rectangle.

This is the answer. 200 + 90 + 0

The place value chart can help us to write a number in expanded notation. When we put 290 into the place value chart, we can recognize that our number is equal to

2 hundreds + 9 tens + 0 units

Image is provided.

17+3 =20

so for every 20 people there are 17 students and 3 faculty

740/20 = 37

37*3 = 111 faculty

37 *17 = 629 students

A 3 x 3 x 4 cuboid can be cut into 3(3) (4) = 36 pieces of 1 x 1 cubes.

Out of the 36 pieces of 1 x 1 cubes, there are no 1 x 1 cube with the four sides painted, there are eight 1 x 1 cubes with simply 3 sides painted, there are twenty 1 x 1 cubes with only 2 sides tinted, there are 6 cubes with only 1 side painted and there are 2 cubes with no side painted.

The chance that the four sides of a randomly selected cube is painted is 0.

The likelihood that only 3 sides of a randomly selected cube is painted is 8 / 36 = 2 / 9.

The chance that only 2 sides of a randomly selected cube is painted is 20 / 36 = 5 / 9.

The likelihood that only 1 side of a randomly selected cube is painted is 6 / 36 = 1 / 6.

The probability that no side of a randomly selected cube is painted is 2 / 36 = 1 / 18.


Thus, the expected value for the painted sides of a randomly selected cube is given by 4(0) + 3(2 / 9) + 2(5 / 9) + 1(1 / 6) + 0( 1 / 18) = 2 / 3 + 10 / 9 + 1 / 6 = 35 / 18 = 1.94

Final answer:

To find the markup on a pair of hiking boots with a cost of $80 and a 15% markup rate, multiply the cost by 0.15 to get a $12 markup. The selling price would then be $80 + $12, totaling $92.

Explanation:

To calculate the markup for a pair of hiking boots that the store buys for $80 with a markup rate of 15%, first convert the percentage to a decimal by dividing by 100.

Markup rate as a decimal: 0.15

Next, multiply the cost price by the decimal.

Markup amount = Cost price × Markup rate

Markup amount = $80 × 0.15

Markup amount = $12

So, the markup on the hiking boots is $12. To find the selling price, you add the markup to the cost price:

Selling price = Cost price + Markup amount

Selling price = $80 + $12

Selling price = $92

The store would sell the hiking boots for $92.

Answer: the answer to this question is

choice c. 17

The answer is B. 40/1 dollars per person; the cost is $40 for each person.

A. In this case, they can be arranged at any order. Since there are 7 people all in all, therefore the number of arrangements is:

number of arrangements = 7P7 = 5040 ways

B. In this case, only the 5 people can be arranged in any order therefore 5P5. However the groom and bride can be interchanged on 2 places, therefore:

number of arrangements = 5P5 * 2 = 240 ways

C. In this case, only the 5 people can be arranged in any order while the groom and bride can no longer be interchanged, therefore:

number of arrangements = 5P5 = 120 ways

Final answer:

The arrangements for the wedding reception receiving line can be calculated using permutations: a) Any order will do, giving 7!, b) The bride and groom must be the last two in line, giving 5!, c) The groom must be last in line with the bride next to him, giving 6! x 2.

Explanation:

The question asks about the number of ways a bride and groom along with five attendants can be arranged in a receiving line at a wedding reception under different conditions. This is a permutations problem where we calculate the arrangements using factorial notation.

Case a: Any order will do.

In this case, we have a total of 7 people (bride, groom, and five attendants) who can be arranged in any order. The number of arrangements is the factorial of 7, which is calculated as 7! (7 factorial). So, the number of ways they can be arranged is 7! (which is 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040 ways).

Case b: The bride and groom must be the last two in line.

Since the bride and groom's positions are fixed at the end of the line, we only need to arrange the five attendants. This is equivalent to 5! arrangements for the attendants. Thus, the number of ways they can be arranged is 5! (which is 5 x 4 x 3 x 2 x 1 = 120 ways).

Case c: The groom must be last in line with the bride next to him.

Here, the bride and groom are fixed in the last two positions, but the bride must be next to the groom. We consider the bride and groom as a single unit in addition to the five attendants. This gives us 6! arrangements for the attendants and the bride-groom unit, multiplied by 2 because the bride and groom can switch places within their unit. So, the number of ways they can be arranged is 6! x 2 (which is 6 x 5 x 4 x 3 x 2 x 1 x 2 = 1,440 ways).

Little Lake Rock is 1811.5 feet long, and Lake Big Rock is 3461.5 feet long.

Let's use algebra to solve this problem.

Let L be the length of Little Lake Rock in feet, and let B be the length of Lake Big Rock in feet.

We are given two pieces of information:

Lake Big Rock is 1650 ft longer than Little Lake Rock: B = L + 1650.

Together, they are 5273 ft long: B + L = 5273.

Now, we have a system of two equations with two variables:

B = L + 1650

B + L = 5273

We can use the first equation to express B in terms of L and then substitute this expression into the second equation:

B = L + 1650

Now, substitute B in the second equation:

(L + 1650) + L = 5273

Combine like terms:

2L + 1650 = 5273

Subtract 1650 from both sides:

2L = 5273 - 1650

2L = 3623

Now, divide by 2 to find the length of Little Lake Rock (L):

L = 3623 / 2

L = 1811.5 feet

Now that we know the length of Little Lake Rock, we can find the length of Lake Big Rock using the first equation:

B = L + 1650

B = 1811.5 + 1650

B = 3461.5 feet

So, Little Lake Rock is 1811.5 feet long, and Lake Big Rock is 3461.5 feet long.

for such more question on length

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Answer:

The bird is at a height of 1.95 km from the ground.

Step-by-step explanation:

It is given that the distance between susan and mark is 7 km and bird is between  susan and mark. The angles of elevation they make are 20º and 50º, respectively.

Draw a perpendicular line from the bird on base.

Let the distance of susan from and the altitude be x.

In triangle ABS,

[tex]tan(20^{\circ})=\frac{AB}{x}[/tex]

[tex]tan(20^{\circ})x=AB[/tex]

[tex]0.364x=AB[/tex]       ..... (1)

In triangle ABM,

[tex]tan(50^{\circ})=\frac{AB}{7-x}[/tex]

[tex]1.192(7-x)=AB[/tex]     ..... (2)

From (1) and (2), we get

[tex]0.364x=1.192(7-x)[/tex]

[tex]1.556x=8342[/tex]

[tex]x=5.36[/tex]

The length of AB is,

[tex]AB=0.364(5.36)=1.95147\approx 1.95[/tex]

Therefore, the he bird is at a height of 1.95 km from the ground.