The average human heart beats 1.15 \cdot 10^51.15⋅10 5 1, point, 15, dot, 10, start superscript, 5, end superscript times per da
2 answers:
Answer:
Step-by-step explanation:
We are told that the average human heart beats times per day and there are
days in one year.
To find number of heart beats in one year we will multiply number of heart beats in one day by number of days in one year.
Now we will solve this problem using exponent properties.
Our answer is in scientific notation we can represent it in standard form as times.
Therefore, average human heart beats in one year or 41975000 times.
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Answer:
a)0,45119
b)1
Step-by-step explanation:
For part A of the problem we must first find the probability that both people in the couple have the same birthday (April 30)
Now the poisson approximation is used
λ=nP=80000*1/133225=0,6
Now, let X be the number of couples that birth April 30
P(X ≥ 1) =
1 − P(X = 0) =
P(X ≥ 1) = 0,45119
B) Now want to find the
probability that both partners celebrated their birthday on th, assuming that the year is 52 weeks and therefore 52 thursday
Now the poisson approximation is used
λ=nP=80000*52/133225=31.225
Now, let X be the number of couples that birth same day
P(X ≥ 1) =
1 − P(X = 0) =
P(X ≥ 1) = 1
Answer:
Step-by-step explanation:
Hello!
To test the claim that eating a healthy breakfast improves the performance of students on their test a math teacher randomly asked 46 students what did they have for breakfast before they took the final exam and classified them as:
<u>Group 1</u>: Ate healthy breakfast
X₁: Number of students that ate a healthy breakfast before the exam and earned 80% or higher.
n₁= 26
<u>Group 2: </u>Did not eat healthy breakfast
X₂: Number of students that did not eat a healthy breakfast before the exam and earned 80% or higher.
n₂= 20
After the test she counted the number of students that got 80% or more in the test for each group obtaining the following sample proportions:
p'₁= 0.50
p'₂= 0.40
The parameters of study are the population proportions, if the claim is true then p₁ > p₂
And you can determine the hypotheses as
H₀: p₁ ≤ p₂
H₁: p₁ > p₂
α: 0.05
≈N(0;1)
pooled sample proportion:
p-value: 0.2514
The decision rule is:
If p-value ≤ α, reject the null hypothesis.
If p-value > α, do not reject the null hypothesis.
The p-value: 0.2514 is greater than the significance level 0.05, the test is not significant.
At a 5% significance level you can conclude that the population proportion of math students that obtained at least 80% in the test and had a healthy breakfast is equal or less than the population proportion of math students that obtained at least 80% in the test and didn't have a healthy breakfast.
So having a healthy breakfast doesn't seem to improve the grades of students.
I hope this helps!
Answer:
Step-by-step explanation:
Hello!
Given the variables
X: Curfew of middle school students (Categorized Yes/No)
Y: Average grade of middle school students. (Categorized: A, B, C, and D)
The objective is to test if there is an association between both variables, you have to conduct a Chi-Square test of independence.
In the null hypothesis, you state that both variables are independent vs. the alternative hypothesis that the variables are dependent.
The hypotheses for this test are:
H₀: Pij= Pi. * P.j i= (1)Yes, (2)No; j= (1)A, (2)B, (3)C, (4)D
H₁: Te variables are not independent.
α: 0.05
Where
Oij is the observed frequency for the i-row and j-column
Eij is the expected frequency for the i-row and j-column
r= number of categories in the rows
c= number of categories in the columns
p-value: 0.687
The p-value is greater than the significance level, so the decision is to not reject the null hypothesis. So using a 5% significance level, you can conclude that having a curfew and the average grades of middle schoolers are two independent variables.
