The larger number is 13 and smaller number is 4
<em><u>Solution:</u></em>
Let the larger number be "a"
Let the smaller number be "b"
<em><u>Given that larger of two numbers is twice the smaller increased by five</u></em>
larger number = twice the smaller increased by five
larger number = twice the smaller number + 5
a = 2b + 5 --- eqn 1
<em><u>Also given that three times the larger exceeds double the smaller by 31</u></em>
Here "exceeds" denotes addition and "times" denotes multiplication
"double" denotes being multiplied by 2
three times the larger number = double the smaller number + 31
3a = 2b + 31 ---- eqn 2
<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "b"</u></em>
Substitute eqn 1 in eqn 2
3(2b + 5) = 2b + 31
6b + 15 = 2b + 31
6b - 2b = 31 - 15
4b = 16
<h3>b = 4</h3>
From eqn 1,
a = 2b + 5
a = 2(4) + 5
a = 8 + 5
<h3>a = 13</h3>
Thus the larger number is 13 and smaller number is 4
Answer:
Falso.
Step-by-step explanation:
Acá tenemos la proposición:
"En una multiplicación, si un factor es un número natural y el otro es un número entero negativo, el producto es siempre menor que cada uno de los factores."
Primero tratemos de demostrar que esto es falso, para ello debemos encontrar un solo ejemplo en el que la proposición sea falsa.
Elijamos al número 1 como el número natural,
Elijamos -10 como el número entero negativo.
El producto es:
1*-10 = -10
Ahora veamos si el producto es menor que cada uno de los factores.
-10 < 1 ?
Si, -10 es menor que 1.
Ahora veamos con el otro factor:
-10 < - 10?
No, un número no puede ser menor que si mismo.
Entonces el producto no siempre es menor que cada uno de los factores.
Entonces la proposición es falsa.
75 is my answer and this is a guess so if its right tell me
Answer:
5≤u
Step-by-step explanation:
Put the words and numbers into an equation:
-5u+27≤2
Put all like terms on one side:
27-2≤5u ---> Do you see what I did there? Now you don't have to work with negatives. :)
Simplify:
25≤5u
Solve:
5≤u
:)
Answer:
The 98% confidence interval for the population proportion of disks which are defective is (0.082, 0.118).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
Suppose a sample of 1536 floppy disks is drawn. Of these disks, 1383 were not defective.
1536 - 1383 = 153
This means that 
98% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 98% confidence interval for the population proportion of disks which are defective is (0.082, 0.118).
