Option B:
Both n and m must be rational.
Solution:
Given information:
Sum of two numbers, n and m are rational.
<u>To find which statements are true:</u>
Option A: Both n and m may be rational but do not have to be.
It is not true because n and m given is rational.
It have to be rational.
Option B: Both n and m must be rational.
Yes, n and m must be rational then only the sum of numbers are rational.
It is true.
Option C: Both n and m must be irrational.
Sum of irrationals will be sometimes irrational and sometimes can't add.
So it is not true.
Option D: One number is rational and the other is irrational.
Rational and irrational cannot be add.
So it is not true.
Option B is true.
Both n and m must be rational.
Answer:
the thrid one
Step-by-step explanation:
Answer:
La respuesta es falso.
Step-by-step explanation:
La respuesta es falso.
Cuando se suman fraccciones con igual denominador, se suman los numeradores (numerador con numerador) y se deja el mismo denominador (el cual es común en ambos). Por ejemplo, la suma de 1/5 + 3/5 da como resultado:
En el caso de fracciones con diferentes denominadores, tampoco se suma numerador con numerador y denominador con denominador. En ese caso se debe encontrar el mínimo común múltiplo.
Por lo tanto, la respuesta es falso.
Espero que te sea de utilidad!
Answer:
The claim is false.
Step-by-step explanation:
Given the data :
13 24 21 37 15 25 18 22 40 32
The sample mean and standard deviation can ben calculated for the given sample.
Using calculator :
Sample mean, xbar = 24.7
Sample standard deviation, s = 9.04
Sample size, n = 10
The hypothesis :
H0 : μ = 17
H1 : μ ≠ 17
The test statistic :
(xbar - μ) ÷ (s/√(n))
(24.7 - 17) ÷ (9.04/√(10))
7.7 / 2.8586990
Test statistic = 2.694
We can obtain the Pvalue, at α = 0.05 ; df = n-1 = 9
Pvalue = 0.0246
Since Pvalue < α ; we reject the null ; Hence, there is significant evidence to conclude that an adult American does not spend average of 17 hours in leisure
4 because 2cups of 1/2 cups=1cup
2x2=4
