Ask question

Ask question

All categories

Brilliant_brown [7]

2 years ago

10

Jesus tiene un terreno cuadrado y su hermano uno rectángulalar,El Largo Del Terreno De Su hermano mide 5 metros más que el lado

que el terreno que Jesús.mientras que el ancho del terreno de Si hermano mide lo mismo que el ancho del terreno cuadrado. ¿Cual es el area total de los terrenos?

1 answer:

oee [108]2 years ago

4 0

Answer:

El terreno de Jesus es cuadrado.

El terreno del hermano es rectangular.

Definimos las variables:

Lh = Largo del terreno del hermano.

Ah = Ancho del terreno del hermano

Lj = Largo del terreno de Jesus.

Sabemos que:

Lh = 5m + Lj

Ah = Lj.

Sabemos que el area de un quadrado de lado L es igual a L^2.

Entonces el area del terreno de Jesus es Lj^2.

El area de un rectangulo de largo L y ancho A es igual a A*L.

El area del terreno del hermano de Jesus es:

Lh*Ah = (5m + Lj)*Lj

= 5m*Lj + Lj^2.

You might be interested in

For a display, identical cubic boxes are stacked in square layers. Each layer consists of cubic boxes arranged in rows that form

serg [7]

Answer:

285 boxes are in the display

Step-by-step explanation:

Given data

top layer box = 1

last row box = 81

to find out

how many box

solution

we know that every row is a square so that if the bottom layer has 81 squares it mean this is 9² and every row has one lesser box

so that next row will have 8^2 and than 7² and so on till 1²

so we can say that cubes in the rows as that

Sum of all Squares = 9² + 8² +..........+ 1²

Sum of Squares positive Consecutive Integers formula are

Sum of Squares of Consecutive Integers = (1/6)(n)(n+1)(2n+1)

here n = 9 so equation will be

Sum of Squares of Consecutive Integers = (1/6) × (9) × (9+1) × (2×9+1)

Sum of Squares of Consecutive Integers = 285

so 285 boxes are in the display

7 0

2 years ago

Jake, Daniel, Timmy, Amy, and Pam are going on a fishing trip. They are taking a pick-up truck which only holds 3 people inside

VashaNatasha [74]

Answer:

It has to be Timmy and it is 412.

Step-by-step explanation:

Uh I did it in my head and it is really hard to explain.

3 0

2 years ago

Read 2 more answers

What is the result of substituting for y in the bottom equation?

zhenek [66]

Answer:

(A) $x + 3= x^2 + 2x-4$

Step-by-step explanation:

Given the equations:

$y = x + 3\\y= x^2 + 2x-4$

Substitution simply means replacing the variable y in the second equation with its equivalent x+3 from the first equation.

Substitution of y into $y=x^2 + 2x-4$ gives us:

$x + 3= x^2 + 2x-4$

The correct option is A.

6 0

2 years ago

A group of 10 students participate in chess club, karate club, or neither.

Afina-wow [57]

Total number of students = 10

As we have to find

P(A/B) = Probability( A when B has happend)

P(A/B)= P(A intersection B)/P(B)

According to given figure only yolanda and Rob are in both club

Therefore,P(A intersection B) $=\frac{2}{10}$

Number of student in karate club =6

P(B) $= \frac{6}{10}$

P(A/B) $= \frac{2}{10}\div{\frac{6}{10}$

Converting division into multiplication by reciprocating the term after division

P(A/B) $= \frac{2}{10}\times{\frac{10}{6}$

On solving we get ,

P(A/B) $= \frac{2}{6}$

P(A/B) $=\frac{1}{3}$

P(A/B) $=0.33$

8 1

2 years ago

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

Gre4nikov [31]

Answer:

A.the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. β = 0.0122

C. β = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o: \mu = 100}$

$\mathtt{H_1: \mu \neq 100}$

A. If the acceptance region is defined as $98.5 < \overline x > 101.5$ , find the type I error probability $\alpha$ .

Assuming the critical region lies within $\overline x < 98.5$ or $\overline x > 101.5$ , for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is $\mu = 100$

∴

$\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}$

$\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}$

when $\mu = 100$

$\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }$

$\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }$

$\mathtt{\alpha = P ( Z 2.25) }$

$\mathtt{\alpha = P ( Z$

From the standard normal distribution tables

$\mathtt{\alpha = 0.0122+( 1- 0.9878) })$

$\mathtt{\alpha = 0.0122+( 0.0122) })$

$\mathbf{\alpha = 0.0244 }$

Thus, the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis $\mathtt{H_o}$

Thus;

β = P( type II error) - P( fail to reject $\mathtt{H_o}$ )

∴

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 103$

$\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }$

$\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}$

From standard normal distribution table

β = 0.0122 - 0.0000

β = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 105$

$\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }$

$\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}$

From standard normal distribution table

β = 0.0000 - 0.0000

β = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0

2 years ago