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3 years ago

Please help me with this!

Mathematics

2 answers:

kotegsom [21]3 years ago

7 0

If the figure has rotational symmetry, the angle of rotation about the center that results in an image that matches the original figure would be 120 degrees.

-Hiadamcom from Agarvated

Fed [463]3 years ago

3 0

1. B

2. A

3. B

4. B

5. C

6. C

7. B

8. D

9. A

10. A

100% Correct

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I need help on finding the area of a parallelogram

Hoochie [10]

Answer:

1. 540 sq. ft. 2. 7,200 sq. m. 3. 276 sq. yd. 4. 6,000 sq. mm.

Step-by-step explanation:

You find the area of a parallelogram by doing the same as you would a rectangle, multiplying length by width. (If you imagine slicing off the triangle shape on the end of all of these and putting it on the other side, it will always make a rectangle.)

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3 years ago

Write a subtraction problem that involves two positive integers and has a negative answer

Phantasy [73]

Answer:

15-16=-1

Step-by-step explanation:

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3 years ago

Pleasee help ASAP!! No links pleaseee!!!

Ilya [14]

Given:

The vertices of the rectangle ABCD are A(0,1), B(2,4), C(6,0), D(4,-3).

To find:

The area of the rectangle.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using the distance formula, we get

$AB=\sqrt{(2-0)^2+(4-1)^2}$

$AB=\sqrt{(2)^2+(3)^2}$

$AB=\sqrt{4+9}$

$AB=\sqrt{13}$

Similarly,

$BC=\sqrt{(6-2)^2+(0-4)^2}$

$BC=\sqrt{(4)^2+(-4)^2}$

$BC=\sqrt{16+16}$

$BC=\sqrt{32}$

$BC=4\sqrt{2}$

Now, the length of the rectangle is $AB=\sqrt{13}$ and the width of the rectangle is $BC=4\sqrt{2}$ . So, the area of the rectangle is:

$A=length \times width$

$A=\sqrt{13}\times 4\sqrt{2}$

$A=4\sqrt{26}$

$A\approx 20$

Therefore, the area of the rectangle is 20 square units.

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3 years ago

The graph of a non-invertible function passes through the points (- 1, 2), (0, 6) and (3, - 4) How is the inverse of this functi

KATRIN_1 [288]

Answer:

The order of ordered pairs of a function and its inverse reverse. The graph of the inverse of this function passes through the points (2,-1), (6,0), (-4,3).

Explanation:

I just took the test.

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3 years ago

Of the entering class at a college, % attended public high school, % attended private high school, and % were home schoole

Veronika [31]

Answer:

(a) The probability that the student made the Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

A = a student attended public high school

B = a student attended private high school

C = a student was home schooled

D = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

$P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})$

Compute the probability that the student made the Dean's list as follows:

$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$

$=(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655$

Thus, the probability that the student made the Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D)}$

$=\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411$

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

$P(C^{c}|D^{c})=1-P(C|D^{c})$

$=1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185$

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0

2 years ago