Write 10x7 – 9x2 + 15x3 + 9 in standard form. Question 5 options:
2 answers:
Answer:
Option B.
Step-by-step explanation:
The given polynomial is
In standard form of a polynomial, the terms are arranged according to its degree from left to right. It means term with highest degree should be on the left side.
Degree of are 7, 2, 3 and 0.
Arrange the terms according to its degree.
Hence, the correct option is B.
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Answer:
a) n(none) = 25
b) n(PE but not Bio) = 25
c) n(ENG but not both BIO and PE) = 55
d) n(students that did not take Eng or Bio) = 40
e) P( Students did not take exactly two subjects) = 0.65
Step-by-step explanation:
From the Venn diagram drawn:
a) Number of students that took none
n(Freshmen) = 135
n(all three) = 5
n (PE and Bio) = 10
n(PE and Eng) = 15
n(Bio and Eng) = 7
n (PE and Bio only) = 10 - 5 = 5
n(PE and Eng only) = 15 - 5 = 10
n(Bio and Eng only) = 7 - 5 = 2
n(PE only) = 35 - 5 - 5 - 10 = 15
n(Bio only) = 42 - 5 - 5 - 2 = 30
n(Eng only) = 60 - 10 - 5 -2 = 43
n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)
135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)
135 = 110 + n(none)
n(none) = 135 - 110
n(none) = 25
b)Number of students that too PE but not Bio
n(PE but not bio)= n(PE only) + n(PE and Eng only)
n(PE but not Bio) = 15 + 10
n(PE but not Bio) = 25
c) Number of students that took ENG but not both BIO and PE
n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10
n(ENG but not both BIO and PE) = 55
d) Number of students that did not take ENG or BIO
n( students that did not take Eng or Bio) = n(PE only) + n(none)
n(students that did not take Eng or Bio) = 15 + 25
n(students that did not take Eng or Bio) = 40
e) Probability that a randomly-chosen student from this group did not take exactly two subjects
n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)
n( Students that did not take exactly two subjects) = 15 + 30 + 43
n( Students that did not take exactly two subjects) = 88
P( Students did not take exactly two subjects) = 88/135
P( Students did not take exactly two subjects) = 0.65
Using the normal distribution and the central limit theorem, it is found that there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.
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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.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
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- Mean of 700g means that
- Standard deviation of 21g means that
- Sample of 64, thus
- <u>For the sampling distribution of the sample mean</u>, the standard deviation is of
The probability of finding a sample mean mass of 695g or below is the p-value of Z when X = 695, thus:
By the Central Limit Theorem
has a p-value of 0.0284.
0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.
A similar problem is given at brainly.com/question/22934264