Approximate 278 to the nearest tenth.
1 answer:
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The probability of choosing two yellow marbles = 5/26
Step-by-step explanation:
Step 1 :
Given
Total number of marbles in the bag = 13 marbles
Number of yellow marbles = 6.
We choose 2 marbles from the bag without replacing and need to determine the probability that both are yellow.
Step 2 :
Probability of finding yellow marble in the first try = Number of yellow marbles / Total number of marbles =
Once the yellow marble is selected we have 5 yellow marbles among 12 remaining marbles in the bag.
Probability of finding yellow marble in the second try = Number of yellow marbles / Total number of marbles =
Step 3:
P(yellow in first 2 tries) = P(yellow in first try) * P(yellow in second try)
P(yellow in first 2 tries) = =
Step 4:
Answer:
The probability of choosing two yellow marbles = 5/26
Answer:
Jessie scored higher than Reagan.
Step-by-step explanation:
We are given that Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 and standard deviation 100.
Jessie scored 30 on the ACT. The distribution of ACT scores in a reference population is normally distributed with mean 17 and standard deviation 5.
For finding who performed better on the standardized exams, we have to calculate the z-scores for both people.
1. <u>Finding z-score for Reagan;</u>
Let X = distribution of SAT scores
SO, X ~ Normal()
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean = 1000
= standard deviation = 100
Now, Reagan scored 1140 on the SAT, that is;
z-score = = 1.4
2. <u>Finding z-score for Jessie;</u>
Let X = distribution of ACT scores
SO, X ~ Normal()
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean = 17
= standard deviation = 5
Now, Jessie scored 30 on the ACT, that is;
z-score = = 2.6
This means that Jessie scored higher than Reagan because Jessie's standardized score was 2.6, which is 2.6 standard deviations above the mean and Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean.