All categories

3 years ago

LMNP is a rectangle. Find the value of x and the length of each diagonal.

Mathematics

1 answer:

Marta_Voda [28]3 years ago

8 0

Answer:

12.571 length

x= 1.8571 or 13/7

Step-by-step explanation:

You might be interested in

How do i solve <br> 3n= 3.6

quester [9]

Answer:

n = 1.2

Step-by-step explanation:

3n = 3.6

Divide both sides by 3 to get n by itself.

3.6/3 = 1.2

n = 1.2

5 0

3 years ago

Read 2 more answers

Which expression is equivalent to 32

Nadusha1986 [10]

2 times (9+7) is correct.

9 + 7 = 16.
16 multiplied by 2 is 32.

3 0

4 years ago

What's the product: (12x+1)(3x-8)

jasenka [17]

Answer:

the answer would 12

Step-by-step explanation:

5 0

3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago

QUESTION 29

Shalnov [3]

The sum of the set of x - values given to us is calculated as; ∑X = -1

We are given;

X1 = 2, X2 = -8, X3 = 4, X4 = -8 and Y1 = -3, Y2 = -8, Y3 = 10, Y4 = 6

1) ∑X = X1 + X2 + X3 + X4

∑X = 2 + (-8) + 4 + 1 = -1

2) ∑Y = Y1 + Y2 + Y3 + Y4

∑Y = 3 + (-8) + 10 + 6

∑Y = -11

3) ∑ X² = X₁² + X₂² + X₃² + X₄²

∑X² = (2)² + (-8)² + (4)² + (1)²

∑X² = 85

4) ∑Y² = Y₁² + Y₂² + Y₃² + Y₄²

∑X² = (-3)² + (-8)² + (10)² + (6)²

∑X² = 209

Complete question is;

Two variables, X and Y assume the values X1 = 2, X2 = -5, X3 = 4, X4 = 1 and Y1 = -3, Y2 = -8, Y3 = 10, Y4 = 6, respectively. Calculate (a) ∑X, (b) ∑ Y, (c) ∑ X² (d) ∑ Y²

Read more about Sum of data at; brainly.com/question/15858152

#SPJ1

6 0

2 years ago