Response:
a. 55 cars
b. 25 cars
Detailed explanation:
Let’s denote the quantity of cars with stereo systems as N(ss), those with air conditioning as N(ac), and those with sunroofs as N(sr).
We find that:
N(ss) = 30
N(ac) = 30
N(sr) = 40
N(ss and ac and sr) = 15
N(at least two) = 30
a.
To calculate how many cars possess at least one feature (N(at least one) or N(ss or ac or sr)), we apply:
N(ss or ac or sr) = N(ss) + N(ac) + N(sr) - N(ss and ac) - N(ss and sr) - N(ac and sr) + N(ss and ac and sr)
N(ss or ac or sr) = 30 + 30 + 40 - (N(at least two) + 2*N(ss and ac and sr)) + 15
Substituting, we find N(ss or ac or sr) = 30 + 30 + 40 - (30 + 2*15) + 15 = 55
b.
For those cars that have exactly one feature, we have:
N(only one) = N(at least one) - N(at least two)
N(only one) = 55 - 30 = 25
It is classified as a real number.
Answer:
Expiration Date: 1/17/2017
Expiration Time: 4:00am
Preparation Date: 12/3/2016
Preparation Time: 4:00am
Initial Usage Date: 12/7/2016
Detailed Breakdown:
An illustrative depiction of the question has been provided in an image format for clarity.
From the information given, it is noted that her store order arrived on 12/3/2016 at 4am, confirming that both the prep date and time are 12/3/2016 and 4am respectively. The product has a printed expiration date of 1/17/2017, logically indicating that its expiration time is also 4am, in line with the prep time; adding 24 hours leads us back to the same time on the expiration date. Furthermore, we were informed that she utilized the product on 12/7/2016, which marks the initial use date. Based on this information, we can summarize as follows:
Expiration Date: 1/17/2017
Expiration Time: 4:00am
Preparation Date: 12/3/2016
Preparation Time: 4:00am
Initial Usage Date: 12/7/2016
The question is missing some information. It should be phrased as follows:
<span><span>A container has 50 electronic components, with 10 identified as defective. If 6 components are randomly selected from the container, what is the probability that at least 4 of them are not defective? Additionally, if 8 components are drawn at random from the container, what is the probability that exactly 3 are defective?
</span>Answers
<span>Part 1. 0.02
Part 2. </span></span>0.0375<span><span>
</span>Explanation
Probability denotes the likelihood of an event occurring. It is computed as:
probability = (Number of favorable outcomes)/(Number of total outcomes)
Part 1
When 6 components are chosen, if 4 are confirmed functioning, then 2 must be defective.
P(at least 4 functional) = 4/40</span>× 2/10
= 1/10 × 1/5
= 1/50
= 0.02
Part 2
Choosing 8 components, if 3 are defective, then 5 are functioning.
P(3 defective) = 3/40 × 5/10
= 15/400
= 3/80
= 0.0375
Begin by grouping the first two terms together and the last two together, then factor and redistribute
(4x^3+x^2)+(-8x-2)
(x^2)(4x+1)+(-2)(4x+1)
x^2(4x+1)-2(4x+1)
the answer is the first one
