I choose a card at random from a well-shuffled deck of 52 cards. There is a 1/4 probability that the card chosen is a spade, a 1/4 probability that the card is a heart, a 1/4 probability that the card is a diamond, and a 1/4 probability that the card is a club. Both spades and clubs are black cards, whereas hearts and diamonds are red. The events card is a heart and card is a club are

The answer is "disjoint events".

Step-by-step explanation:

Please find the complete question in the attached file.

In this question, the Disjointed events were not connected, throughout this, Some other approach to disorganized events was its lack of common outcomes. When it can come at the same time, two events are exclusive to each other. A whole term, that implies excluding each other is disjoined. If had some are united, the chance of them both happening on one is 0, throughout this question, both heart and club can't be cards at the very same time.

assuming you mean addition: addition and subtraction are reverse operations.

step-by-step explanation:

d

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.