Find the Area and Perimeter of both figures!

Naddik [55]

Answer:

Step-by-step explanation:

a) To find the length of diagonal XT, we can use the distance formula to get $\sqrt{(-3-5)^{2}+(4-(-2))^{2}}=10$

Since this is a rectangle, XT=YW, meaning YW=10 as well.

b) The area of a rectangle is given by length times width. The length is |-3-5|=8, and the width is |-2-4|=6. So the area is (8)(6)=48.

c) The perimeter is just 2(8+6)=28.

8 0

2 years ago

A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o

balu736 [363]

Answer:

a

$P(X \ge 1) = 0.509$

b

$P(X \ge 1) = 0.6807$

Step-by-step explanation:

From the question we are told that

The number of students in the class is N = 20 (This is the population )

The number of student that will cheat is k = 3

The number of students that he is focused on is n = 4

Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally probability mass function is mathematically represented as

$P(X = x) = \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}$

Here C stands for combination , hence we will be making use of the combination functionality in our calculators

Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

Here

$P(X \le 0) = \frac{ ^{3} C_0 * ^{20 - 3} C_{4- 0}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ ^{3} C_0 * ^{17} C_{4}}{ ^{20}C_4}$

$P(X \le 0) = \frac{ 1 * 2380}{ 4845}$

$P(X \le 0) = 0.491$

Hence

$P(X \ge 1) = 1 - 0.491$

$P(X \ge 1) = 0.509$

Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

$P(X \ge 1) = 1 - P(X \le 0)$

$P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]$

Here n = 6

So

$P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]$

$P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]$

$P(X \ge 1) =1- [ \frac{1 * 12376}{38760}]$

$P(X \ge 1) =1- 0.3193$

$P(X \ge 1) = 0.6807$

5 0

3 years ago

What will be the result if the y-intercept of the line shown is changed to an 8, but the slope remains the same? Select one opti

kkurt [141]

Answer:2 down

Step-by-step explanation:

y-intercept is on 10 so if it said that it changed to 8 u subtract 2 which means it goes down

3 0

2 years ago

Plz help look at the Picture thanks so much for the Help

Kisachek [45]

The amount of the markup is 20% of 25.15

EQUATION 1: 20/100 * 25.15

ANSWER TO EQUATION 1: 503/100

Then we have divide 503 and 100

EQUATION 2: 503/100

FINAL ANSWER: $5.03

The amount of the markup is $5.03

3 0

3 years ago

What is the value of 2x3 + 4x2 - 3x2 - 6x when x = 3?

never [62]

Answer:

-10

Step-by-step explanation:

If the equation is 2 x 3+4 x 2-3 x 2- 6(x)= . Than the equation would turn into 2 x 3+4 x 2-3 x 2- 6(3) which equals -10.

6 0

3 years ago