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mylen [45]
2 years ago
5

A teacher listed 28,30,32 and 36 as ages of the students in his class with frequencies 8,10,5 and 7 Respect Respectively.......

A:how many students were in the class ? B: what was the average age of the class? C: what was the range for the students? D:what was the modal age?​
Mathematics
2 answers:
Kamila [148]2 years ago
7 0

Answer:

A: There are 30 students in the class.

B: The average age of the class is 31.5.

C: The range goes from age 28 to age 36.

D: The modal age is 30.

Step-by-step explanation:

A: 8+10+5+7 = 30.

B: 8 + 30 + 32 + 36 = 126

    126/4 = 31.5

C: Lowest age is 28 and highest is 36, making that the range for students' ages.

D: The most frequently appearing age in the study is 30, which makes 30 the modal age.

Margaret [11]2 years ago
6 0

gawd daym that's a lot of students  

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3 years ago
Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus
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Hi! 

To compare this two sets of data, you need to use a t-student test:

You have the following data:

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-Wednesday n2=20;  </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

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5 0
3 years ago
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Shkiper50 [21]

Answer:

C or 8

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3 years ago
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4/9 divided by what equals 12
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2 years ago
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