Jackie has 1/3 of a Hershey bar.steven has 4/12 of a Hershey bar.How much do they have together?
2 answers:
Answer: 2/3
Step-by-step explanation: To find how much of the candy bar they each have, we will have to add the fractions. To add unlike fractions such as 1/3 and 4/12, we first need to find a common denominator. The common denominator of 3 and 12 is simply the least common multiple of 3 and 12 which is 12. To get 12 in the denominator of 1/3, we multiply the numerator and denominator by 4 to get 4/12. Notice that our second fraction already has 12 in the denominator so we can now add these fractions.
+
=
.
Notice that 8/12 can be reduced by dividing both the numerator and denominator by 4.
8 ÷ 4 = 2
12 ÷ 4 = 3
Therefore, Jackie and Steven have 2/3 of a Hershey bar.
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Answer:
2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So
has a pvalue of 0.7881.
1 - 0.7881 = 0.2119
So 21.19% of the students use the computer for longer than 40 minutes.
Out of 10000
0.2119*10000 = 2119
2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.