12(2)+5
Dang, you're an elementary school er? Good grief i'm and 8th grade math one student!
Answer:
$9.10
Step-by-step explanation:

455 ÷ 50 = 9.1. The student tickets cost $9.10. (I think)
Answer:
B
Step-by-step explanation:
Hope it helps
Using the normal distribution, it is found that 0.26% of the items will either weigh less than 87 grams or more than 93 grams.
In a <em>normal distribution</em> with mean
and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
In this problem:
- The mean is of 90 grams, hence
.
- The standard deviation is of 1 gram, hence
.
We want to find the probability of an item <u>differing more than 3 grams from the mean</u>, hence:



The probability is P(|Z| > 3), which is 2 multiplied by the p-value of Z = -3.
- Looking at the z-table, Z = -3 has a p-value of 0.0013.
2 x 0.0013 = 0.0026
0.0026 x 100% = 0.26%
0.26% of the items will either weigh less than 87 grams or more than 93 grams.
For more on the normal distribution, you can check brainly.com/question/24663213
Answer:
1/6
Step-by-step explanation:
So we are subtracting a negative which is the same as adding. Our equation becomes -1/3+1/2. To add fractions we need a common denominator (the denominator of a fraction is the number on the bottom) To fine the least common denominator we need to find the lowest number that both 3 and 2 go into which is 6. Then we will multiply each fraction by the number that will give us 6 for a denominator which is 2 for -1/3 and 3 for 1/2. So -1/3 times 2 is -2/6 and 1/2 times 3 is 3/6. Our equation is now -2/6+3/6. Now we will add the numerators ( the numerator is the number on the top of the fraction) and the denominators stay the same when adding or subtraction fractions -2/6+3/6=1/6
I hope this helps and please let me know if there is anything you are confused about or is still unclear, I will be happy to help!