(−7) Times (−4) whats the answer
2 answers:
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Answer:
Choice number one:
.
Step-by-step explanation:
- Let
be the event that the number on the first card is even.
- Let
be the event that the number on the second card is even.
The question is asking for the possibility that event and
happen at the same time. However, whether
occurs or not will influence the probability of
. In other words,
and
are not independent. The probability that both
and
occur needs to be found as the product of
- the probability that event
- the probability that event
5 out of the ten numbers are even. The probability that event occurs is:
.
In case A occurs, there will only be four cards with even numbers out of the nine cards that are still in the bag. The conditional probability of getting a second card with an even number on it, given that the first card is even, will be:
.
The probability that both and
occurs will be:
.
32.5 % of lab are wireless devices
<em><u>Solution:</u></em>
Given that there was a total of 30 computers and the computer lab is removing 3 broken computers and adding 13 wireless devices
Now we can first find out total number of computers and devices in lab
total number of computers and devices in lab = 30 computers - 3 + 13 wireless devices
total number of computers and devices in lab = 30 - 3 + 13 = 40
So there are 40 computers and wireless devices in lab
<em><u>To find: what percent of the lab are wireless devices</u></em>
So we have to find what percent of 40 is 13
Let "a" be the required percent
a % of 40 = 13
32.5 % of of lab are wireless devices