A____________ is sometimes a rhombus
A. Triangle
B. Square
C. Rectangle
D. Trapezoid

Answer: 1 What???

Step-by-step explanation:

The only plane shape in which every point on the figure is the exact same distance from the center is; The Circle

There are different types of plane shapes such as Square, rectangle, Circle, Trapezoid, Parallelogram, Rhombus, Kite, Triangle e.t.c.

Now, Except for the circle and Triangle, every other listed plane shape above is a quadrilateral because they have 4 sides. Now, no quadrilateral has all its' sides equal from the center. Likewise, no triangle has all its' sides equal from the center.

The only plane shape with all its' sides equal from the center is the circle because the radius is equal all through.

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Answer:

arc BX = central angle BCX

inscribed BNX = 1/2 of central angle BCX

inscribed BNX = 1/2 of arc BX

Step-by-step explanation:

arc BX = central angle BCX

inscribed BNX = 1/2 of central angle BCX

inscribed BNX = 1/2 of arc BX

Answer:

6

Step-by-step explanation:

The ratio of boys to girls is 2:5

let the number of boys be 2x -------(i)

and number of girls be 5x

and the total number of students is 21

therefore we know that..

2x+5x = 21

on solving for x,

7x=21

x= 21/7

x=3

now, by putting the value of x in (i),

2x = 2×3 = 6

therefore, there are 6 boys in the classroom.

you find the no. of girls similarly.

hope it helps !!...

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The circumference of the circle that Jason drew is gotten as; 62.80 inches

Since Jason lifted the pen and pulled the chain as far as it would reach before drawing a circle, then it means that;

radius of circle; r = 10 inches

Formula for circumference is;

C = 2πr

Thus;

C = 2 * 3.14 * 10

C = 62.80 inches

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Answer:

[tex]\boxed{\sf \dfrac{14}{95}}[/tex]

Explanation:

Black ribbons = 8 ribbons

Green ribbons = 12 ribbons

Total ribbons: 12 + 8 = 20 ribbons

[tex]\sf Probability = \dfrac{Favorable \ Outcomes}{Total \ Possible \ Outcomes}[/tex]

For Lila:

[tex]\sf Lila \ picks \ black \ : \ \dfrac{8}{20}[/tex]

Left black ribbons: 8 - 1 = 7 black ribbons

Left total ribbons: 20 -1 = 19 total ribbons

For Jessica:

[tex]\sf Jessica \ picks \ black \ : \ \dfrac{7}{19}[/tex]

Probability that they both pick black ribbons:

[tex]\sf \rightarrow \dfrac{8}{20} \ x \ \dfrac{7}{19}[/tex]

[tex]\sf \rightarrow \dfrac{14}{95}[/tex]

The answer will be option D which is to Assign the homework questions the numbers 0–9. Use a random number table to select the first three unique numbers between 0 and 9.

Randomization allows a sample or small groups with similar characteristics to determine a probability or to hypothesize an event, that is why this type of sample is important in an experiment, that is, to carry it out in a random way.

Given that:-

A teacher assigns 10 questions for homework and wants to randomly pick 3 questions to be graded.

Assign the homework questions the numbers 0–9. Use a random number table to select the first three unique numbers between 0 and 9. These represent the questions that will be graded.

Hence the answer will be option D which is to Assign the homework questions to the numbers 0–9. Use a random number table to select the first three unique numbers between 0 and 9.

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The distribution of y is [tex]P(y) = ^5C_y * (60\%)^y * (1 - 60\%)^{5-y}[/tex]

The given parameters are:

Probability of success, p = 60%

Sample size, n = 5

The distribution of y is represented as:

[tex]P(y) = ^nC_y * p^y * (1 - p)^{n-y}[/tex]

So, we have:

[tex]P(y) = ^5C_y * (60\%)^y * (1 - 60\%)^{5-y}[/tex]

Hence, the distribution of y is [tex]P(y) = ^5C_y * (60\%)^y * (1 - 60\%)^{5-y}[/tex]

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The relative frequency table is:

Class         Frequency  Relative frequency

40 - 50         4                     0.1

50 - 60          6                     0.15

60 - 70           11                     0.18

70 - 80            9                     0.23

80 - 90              7                    0.18

90 - 100             3                   0.08

The relative frequency measures how often a value appears relative to the sum of the total values. The class width is 10. The class width is the difference between the upper and lower boundary (e.g. 100 - 90).

Frequency is the number of times the numbers in the class interval appear in the data set. Relative frequency is the frequency of a class interval divided by the total frequency of the data set.

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Answer:

Step-by-step explanation:

Answer:

i agree with the other anwser!!

Step-by-step explanation:

Answer:

P(king or diamond) = P(king) + P(diamond) - P(king and diamond) P(king or diamond) = (4/52) + (13/52) - (1/52) P(king or diamond) = 16/52 Page 5 P(king or diamond) = 4/13.

Answer:

52

Step-by-step explanation:

Answer:

1. Find the prime factorization of each number.

2. Write each number as a product of primes, matching primes vertically when possible.

3. Bring down the same numbers in each column and the remaining numbers in each column.

4. Multiply the factors to get the LCM.

for example the lcm of 24 and 36

= 24 = (2 × 2 × 2 × 3) = 2^3 × 3^1

 and 36 = (2 × 2 × 3 × 3) = 2^2 × 3^2

LCM(24, 36) = 2^2 × 3 × 2 × = 72

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Expand 13 as 10 + 3.

10 = 2×5, so

13 = 2×5 + 3×1

or

13 = 2×5¹ + 3×5⁰ = 23₅

Answer:

1/2 this is the correct answer to the question

The measure of each angle is 43.25 and 46.75

The attached image represents the missing piece in the question.

From the image, we can see that:

3x - 13 + 5x - 47 = 90 ---- complementary angles

Evaluate the like terms

8x = 150

Divide by 8

x = 18.75

Substitute x = 18.75 in the given angles

3x-13 = 3 * 18,75 - 13 = 43.25

5x -47 = 5 * 18.75 - 47 = 46.75

Hence, the measure of each angle is 43.25 and 46.75

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Since Sally replaces the card after each draw, there are 52 choices for each card that can be drawn.

In a standard deck of cards, there are 4 aces. Then the probability of drawing an ace is 4/52 = 1/13. The same reasoning applies to drawing a queen or a six. Then the probability of drawing an ace, queen, and six (in that particular order) is 1/13³.

But Sally can also draw these 3 specific cards in any order to end up with the same hand. There are 3! = 6 ways of doing so:

AQ6, A6Q, QA6, Q6A, 6AQ, 6QA

That is, 3 choices for the first card; 2 choices for the second; and 1 for the third, so a total of 3 • 2 • 1 = 3! = 6 hands.

Then the probability of getting a hand consisting of an ace, queen, and six is

3! • 1/13³ = 6/2197 ≈ 0.002731

Answer:

4^5

Step-by-step explanation:

When we are multiplying expressions with the same base, we add the exponents

a^b * a^c = a^(b+c)

4^2 * 4^3 = 4^(2+3) = 4^5

Answer:

question is not correct

Step-by-step explanation:

3(x+2)=11+3x+5

3x+6=11+3x+5, once 3x is subjected from both sides it becomes unsolvable

Answer: Infinite solution

Step-by-step explanation: 3x + 6 = 11 + 3x - 5

                                             -3x

6 = 11 - 5

6 = 6

The answer is C.

Step-by-step explanation:

I can't explain but I literally just did this question and I got it right

Answer:

the answer will be 1 7/3 I believe

Answer:

See below ~

Step-by-step explanation:

Perimeter

Area

Answer:

Step-by-step explanation:

9x-2=4x+9

1. 9x-4x-2=4x-4x+9 always isolate a variable which means move 4x to the other side to get 9 by its self.

2. 5x-2+2=9+2

3. 5x=11

4. 5x/5=11/5 get x by itself.

5. x=2.2

check work.

9(2.2)-2=4(2.2)+9

17.8=17.8

The interquartile range = upper quartile - lower quartile = 18 - 15 = 3 (option A).

The interquartile range (IQR) of a data = upper quartile - lower quartile.

In the box-and-whisker plot given:

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