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prisoha [69]
3 years ago
9

At the Grassy Meadows Mennonite School, the total number of students in grade 7-9 is the lowest commom multiple of the numbers o

f students in grades 8 and 9 . Eight grade has 5 students, ninth grade has 3 students. how many students are in seventh grade?
please help
Mathematics
1 answer:
kobusy [5.1K]3 years ago
4 0

Answer:

Seven  

Step-by-step explanation:

Let G7, G8, and G9 be the numbers of students in the respective grades.

Then, G7 + G8 + G9 = LCM of G8 and G9

<em>Data: </em>

G8 = 5

G9 = 3

<em>Calculations: </em>

<em>Multiples of 5</em>: 1 2          10 15

<em>Multiples of 3</em>: 1   3 6 9       15

The <em>LCM</em> of 5 and 3 is 15.

G7 + 5 + 3 = 15     Combine like terms

     G7 + 8 = 15     Subtract 8 from each side

           G7 = 7

There are seven students in Grade 7.


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An NBA fan named Mark claims that there are more fouls called on his team 1 point
inn [45]

Answer:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

df = n-1=34-1=33

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

Step-by-step explanation:

Information provided

\bar X=12.2 represent the sample mean fould against

s=1.6 represent the sample standard deviation

n=34 sample size  

represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is higher than 11.5 fouls per game:  

Null hypothesis:\mu \leq 11.5  

Alternative hypothesis:\mu > 11.5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

The statistic is given by:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

P value

The degreed of freedom are given by:

df = n-1=34-1=33

Since is a one side test the p value would be:  

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

7 0
3 years ago
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