Answer:
An IF statement can be nested within another IF statement to evaluate multiple conditions. Study the formulas in the options below. Which one is correct? a) =IF(A1>=150,"Bonus Paid", IF(A1>=100,"Voucher Given",IF(A1<99,"Further Training"."Error"))) b) =IF(IF(IF(A1>=150,"Bonus Paid", A1>=100,"Voucher Given", A1<99,"Further Training","Error">>> c) =IF(A1>=150,"Bonus Paid"),IF(A1>=100, "Voucher Given"),IF(A1<99,"Further Training","Error"))) d) All of the formulas listed above are incorrect.
8/7/2023, 2:16:06 PM
An IF statement can be nested within another IF statement to evaluate multiple conditions. Study the formulas in the options below. Which one is correct? a) =IF(A1>=150,"Bonus Paid", IF(A1>=100,"Voucher Given",IF(A1<99,"Further Training"."Error"))) b) =IF(IF(IF(A1>=150,"Bonus Paid", A1>=100,"Voucher Given", A1<99,"Further Training","Error">>> c) =IF(A1>=150,"Bonus Paid"),IF(A1>=100, "Voucher Given"),IF(A1<99,"Further Training","Error"))) d) All of the formulas listed above are incorrect.
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8/7/2023, 2:17:08 PM
The correct formula is c) =IF(A1>=150,"Bonus Paid",IF(A1>=100,"Voucher Given",IF(A1<99,"Further Training","Error"))).
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8/7/2023, 2:16:57 PM
In a group of 20 students, 9 are taking history class, 6 are taking math class and 3 are taking both history and math. If a student is randomly selected from the group, what is the probability that he/she will be taking either history, math or both? Identify the events with letters and then calculate the probability,
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8/7/2023, 2:17:08 PM
Let's define the events as follows:
A: Student taking history class
B: Student taking math class
To calculate the probability of a student taking either history, math, or both, we need to add the probabilities of events A and B and then subtract the probability of the intersection of events A and B (students taking both history and math).
The probability of event A (P(A)) is 9/20 since 9 out of 20 students are taking history.
The probability of event B (P(B)) is 6/20 since 6 out of 20 students are taking math.
The probability of the intersection of events A and B (P(A ∩ B)) is 3/20 since 3 out of 20 students are taking both history and math.
Therefore, the probability of a student taking either history, math, or both (P(A ∪ B)) is calculated as follows:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
= 9/20 + 6/20 - 3/20
= 12/20
= 0.6
So, the probability that a randomly selected student will be taking either history, math, or both is 0.6 or 60%.