Pascals triangle to the 6th:
1 x^0
1 1 x^1
1 2 1 x^2
1 3 3 1 x^3
1 4 6 4 1 x^4
1 5 10 10 5 1 x^5
1 6 15 20 15 6 1 x^6<span>
</span>the problem is to the 6th power so your going to use the 6th row of pascals triangle (don't count the first row). these numbers represent the coefficients of the variables
1(d-5y)^6 + 6(d-5y)^5 + 15(d-5y)^4 + 20(d-5y)^3 + 15(d-5y)^2 + 6(d-5y) + 1
then simplify
Your Welcome!!
(Hopefully they're all right)
Answer:
Step-by-step explanation:
Given
See attachment for triangles
Solving (a)
The tan of an angle is:
From the given triangle.
So, we have:
Solving (b)
The cos of an angle is:
From the given triangle.
So, we have:
The probability of choosing a number that is not a multiple of 2 is P = 0.44
<h3 /><h3>How to find the probability?</h3>
We need to count the number of options for each digit.
- For the first digit, we have 8 options {1, 2, 3, 4, 5, 6, 7, 8}
- For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
- For the third digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}.
The total number of combinations is the product between the numbers of options:
C = 8*9*9 = 648
If we want our number to not be a multiple of 2 then it must end in a odd digit, the combinations that meet that condition are:
- For the first digit, we have 8 options {1, 2, 3, 4, 5, 6, 7, 8}
- For the second digit, we have 9 options {0 ,1, 2, 3, 4, 5, 6, 7, 8}
- For the third digit, we have 4 options {1, 3, 5, 7}.
C = 8*9*4 = 288
Then the probability of selecting a 3 digit number that is not a multiple of 2 is:
P = 288/648 = 0.44
If you want to learn more about probability, you can read:
brainly.com/question/251701
Answer:
The probability is 0.508 = 50.8%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with a mean weight of 0.8544 g and a standard deviation of 0.0525 g.
This means that 
If 1 candy is randomly selected, find the probability that it weighs more than 0.8535 g.
This is 1 subtracted by the pvalue of Z when X = 0.8535. So
has a pvalue of 0.492
1 - 0.492 = 0.508
The probability is 0.508 = 50.8%.
