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vaieri [72.5K]
3 years ago
14

There are 9 classes of 25 students each 4 teachers in two times as many chaperones as teachers each bus hold a total of 45 peopl

e what is the least number of busses needed for the field trip​
Mathematics
1 answer:
Tanya [424]3 years ago
6 0

Answer:

Least number of bus require for trip = 5 buses (Approx)

Step-by-step explanation:

Given:

Total number of classes = 9

Number of student in each class = 25

Number of teacher = 4

Number of chaperones = Double of teacher

Bus hold = 45 people

Find:

Least number of bus require for trip

Computation:

Total number of student = 9 × 25

Total number of student = 225

Number of chaperones = 4 × 2

Number of chaperones = 8

Total people = 225 + 8 + 4

Total people = 237

Least number of bus require for trip = Total people / Bus hold

Least number of bus require for trip = 237 / 45

Least number of bus require for trip = 5.266

Least number of bus require for trip = 5 buses (Approx)

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Zappos is an online retailer based in Nevada and employs 1,300 employees. One of their competitors, Amazon, would like to test t
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Answer:

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Step-by-step explanation:

Data given

\bar X=33.9 represent the sample mean

s=4.1 represent the sample standard deviation

n=22 sample size  

\mu_o =26 represent the value that we want to test

\alpha=0.025 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

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Null hypothesis:\mu \geq 36  

Alternative hypothesis:\mu < 36  

The statistic is given by:

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

And replacing we got:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

Now we can calculate the critical value but first we need to find the degreed of freedom:

df = n-1= 22-1=21

So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:

t_{\alpha/2}= -2.08

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

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