Solution: We are given:
Let 
be the weight (oz) of laptop
We have to find 
To find the this probability, we need to find the z score value.
The z score is given below:
Now, we have to find 
Using the standard normal table, we have:
0.9236 or 92.36% of laptops are overweight
Answer:
167/346 or 0.483
Step-by-step explanation:
From the question given above, the following data were obtained:
Number of Tails (T) = 167
Number of Heads (H) = 179
Probability of tail, P(T) =?
Next, we shall determine total outcome. This can be obtained as follow:
Number of Tails (T) = 167
Number of Heads (H) = 179
Total outcome (S) =?
S = T + H
S = 167 + 179
Total outcome (S) = 346
Finally, we shall determine the probability of tails. This can be obtained as follow:
Number of Tails (T) = 167
Total outcome (S) = 346
Probability of tail, P(T) =?
P(T) = T / S
P(T) = 167 / 346
P(T) = 0.483
Thus, the probability of tails is 167/346 or 0.483
<h3>Answer: </h3>
The GCF is 4
The polynomial factors to 
==========================================================
Further explanation:
Ignore the x terms
We're looking for the GCF of 12, 4 and 20
Factor each to their prime factorization. It might help to do a factor tree, but this is optional.
- 12 = 2*2*3
- 4 = 2*2
- 20 = 2*2*5
Each factorization involves "2*2", which means 2*2 = 4 is the GCF here.
We can then factor like so
The distributive property pulls out that common 4. We can verify this by distributing the 4 back in, so we get the original expression back again.
The polynomial inside the parenthesis cannot be factored further. Proof of this can be found by looking at the roots and noticing that they aren't rational numbers (use the quadratic formula).
Answer:
○ -1
Step-by-step explanation:
Looking closely at this piecewise function, when the line on the left-hand side intersects at <em>-</em><em>3</em><em> </em><em>=</em><em> </em><em>y</em><em>,</em><em> </em><em>x</em><em> </em>is -1.
I hope this helps, and as always, I am joyous to assist anyone at any time.
Step-by-step explanation:
m
one set of ordered pair
b
