189 tickets were purchased on Saturday. The ratio of children's tickets to adult tickets is 8:1, indicating that 8 times as many children's tickets were sold compared to adult tickets. Let c represent the number of children's tickets and a the number of adult tickets. Therefore, 8a = a + 147. By subtracting a from both sides, we find 7a = 147. Upon dividing both sides by 7, we find a = 21 adult tickets. By multiplying the number of adult tickets by 8, we discover that 21 * 8 = 168 children's tickets. Adding these together gives a total of 168 + 21 = 189 tickets sold on Saturday.
x y
1 290
2 280
3 270
4 260
5 250
6 240
7 230
8 220
9 210
10 200
11 90
12 180
13 170
14 160
15 150
16 140
17 130
18 120
19 110
20 100
21 90
22 80
23 70
24 60
25 50
26 40
27 30
28 20
29 10
30 0
Al establecer un sistema de ecuaciones lineales, se determina que el total de obreros que contrató el ingeniero civil es 50.
Para descubrir cuántos obreros fueron contratados, se debe formular un sistema de ecuaciones lineales con dos ecuaciones y dos incógnitas, de la siguiente manera:
(1)
(2)
Donde:
M: representa la incógnita correspondiente al capital del ingeniero
x: simboliza la incógnita que denota el total de obreros
De la ecuación (1) podemos deducir:
(3)
Utilizando la ecuación (3) en la ecuación (2) podríamos calcular el valor de x (número de obreros):
El monto inicial del ingeniero, según la ec (3), sería:
Por lo que, el número de obreros contratados por el ingeniero es 50.
Si quieres conocer otro método para resolver sistemas de ecuaciones lineales, puedes ingresar aquí:
Espero que esto te ayude!
We start with the following information:
p = probability = 0.12<span>
n = total number of students = 39 </span>
x = number of left-handers = 5<span>
u = mean = p * n = 4.68
σ = standard deviation = √(n*p*(1-p)) = √(39 * 0.12 * 0.88) =
2.03</span>
Finding the z score:
z = (x – u) / σ
<span>
z = (5 – 4.68) / 2.03
</span>
z
= 0.1576 = 0.16
<span>
</span>Applying standard tables for z gives the p value as:
p value = 0.5636 = 56.36%
Consequently, there is a 56.36% probability.
In certain cases, a function necessitates multiple formulas to achieve the desired outcome. An example is the absolute value function \displaystyle f\left(x\right)=|x|f(x)=∣x∣. This function applies to all real numbers and yields results that are non-negative, defining absolute value as the magnitude or modulus of a real number regardless of its sign. It indicates the distance from zero on the number line, requiring all outputs to be zero or greater.
<pwhen inputting="" a="" non-negative="" value="" the="" output="" remains="" unchanged:="">
\displaystyle f\left(x\right)=x\text{ if }x\ge 0f(x)=x if x≥0
<pwhen inputting="" a="" negative="" value="" the="" output="" is="" inverse:="">
\displaystyle f\left(x\right)=-x\text{ if }x<0f(x)=−x if x<0
Due to the need for two distinct operations, the absolute value function qualifies as a piecewise function: a function defined by several formulas for different sections of its domain.
Piecewise functions help describe scenarios where rules or relationships alter as the input crosses specific "boundaries." Business contexts often demonstrate this, such as when the cost per unit of an item decreases past a certain order quantity. The concept of tax brackets also illustrates piecewise functions. For instance, in a basic tax system where earnings up to $10,000 face a 10% tax, additional income incurs a 20% tax rate. Thus, the total tax on an income S would be 0.1S when \displaystyle {S}\leS≤ $10,000 and 1000 + 0.2 (S – $10,000) when S > $10,000.
</pwhen></pwhen>
