Answer: The correct answer is option C: Both events are equally likely to occur
Step-by-step explanation: For the first experiment, Corrine has a six-sided die, which means there is a total of six possible outcomes altogether. In her experiment, Corrine rolls a number greater than three. The number of events that satisfies this condition in her experiment are the numbers four, five and six (that is, 3 events). Hence the probability can be calculated as follows;
P(>3) = Number of required outcomes/Number of possible outcomes
P(>3) = 3/6
P(>3) = 1/2 or 0.5
Therefore the probability of rolling a number greater than three is 0.5 or 50%.
For the second experiment, Pablo notes heads on the first flip of a coin and then tails on the second flip. for a coin there are two outcomes in total, so the probability of the coin landing on a head is equal to the probability of the coin landing on a tail. Hence the probability can be calculated as follows;
P(Head) = Number of required outcomes/Number of all possible outcomes
P(Head) = 1/2
P(Head) = 0.5
Therefore the probability of landing on a head is 0.5 or 50%. (Note that the probability of landing on a tail is equally 0.5 or 50%)
From these results we can conclude that in both experiments , both events are equally likely to occur.
There are no tens, ones, or tenths.
4,237,300.04
Answer: 1/4
Hope this helps : )
Ok let me start it from here.
you are studying about properties of natural numbers or whole number or may be integers.
So , In this Question you have to follow property of addition of integers.
⇒7+3=( You are at the point 7 on the number line or on the tight rope , you are moving (+3) on the right direction.) So you will reach at 7+1+1+1=10
⇒4+2=( You are at the point 4 on the number line or on the tight rope , you are moving (+2) on the right direction.) So you will reach at 4+1+1=6
In first case Cecil has taken total walk of 10 units and in second case Cecil has taken a walk of 10 units.
C. Both options A and B will allow him to meet his goal.
Looking at Drake's situation after 4 weeks, he only has $470 saved. By
his original plan, he should have had $500 saved. So he's $30 short of
his goal and has 2 weeks until his originally planned class. If he goes
with option A and takes the later class, he will save an additional $125
which is more than enough to make up the $30 short fall. So option A
will work for him to save enough money for his class. With option B, he
will save $140 for the last 2 weeks of his plan giving him a savings of
$280 for the last 2 weeks. Adding the $470 he's already saved will give
him a total savings of $470 + $280 = $750 which is enough for him to
attend his class. So option B will also allow Drake to attend his
desired class. Both options A and B allow him to meet his goal. Hence,
the answer is "c".
