The answer to this question is 22
Answer: 1.5
Step-by-step explanation:Let the standard deviation of the shoe size be represented by ' ' and the mean shoe size be represented by ' = 7.5'. Also, the sizes of the shoe be represented by ' '. So, . Hence, the standard deviation of the shoe size data for the math class is 1.5.
To Find The Answer, We Need To Add All Of Them Together, Then Divide By the Number Of Values. So:
64+23+78+82+91
-----------------------
5
<span>64+23+78+82+91 = 338.
</span>
338
-----
5
338/5 = 67.6
So, The Average Is 67.6
Answer:
X= 5, -9
Step-by-step explanation:
(X+2)² =49
Expand the bracket,
(X+2) (X+2)= 49
Apply the distributive property;
X(X+2) +2 (X+2) =49
X²+2X+2X+4=49
X²+4X+4=49
Move 49 to the left side of the equation;
X²+4X+4-49=0
X²+4X-45=0
Apply Factorisation method;
Consider the form
a²+bx+c=0
Find two numbers whose sum is equal to b and whose product is equal to c.
Comparing with our equation;
B =4 and C =45
We can use 9 and -5, this is because ;
9+(-5)=4 and 9*(-5)= 45.
Replace X +4X-45=0 with (X-5) (X+9)=0
Therefore;
X-5=0
X+9=0
Moving to the left side of the equation;
Therefore X= 5 and -9
Notice that
So as
you have
. Clearly
must converge.
The second sequence requires a bit more work.
The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then
will converge.
Monotonicity is often easier to establish IMO. You can do so by induction. When
, you have
Assume
, i.e. that
. Then for
, you have
which suggests that for all
, you have
, so the sequence is increasing monotonically.
Next, based on the fact that both
and
, a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.
We have
and so on. We're getting an inkling that the explicit closed form for the sequence may be
, but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.
Clearly,
. Let's assume this is the case for
, i.e. that
. Now for
, we have
and so by induction, it follows that
for all
.
Therefore the second sequence must also converge (to 2).