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user100 [1]
3 years ago
12

the line for the dunking machine was twice as long as the cake walk line the line for the Cakewalk was one third of the length o

f the line for the hoop shoot if there were 12 people in line for the hoop shoot how many people were in line for the dunking machine​
Mathematics
1 answer:
Amanda [17]3 years ago
3 0

Answer:

Total number of people in the line for the dunking machine is 8.

Step-by-step explanation:

Here, let us assume the number of people in the line for hoop shoot = 12

So, the number of people in the line for cakewalk

=  One- Third of the number of people in line for Hoop Shoot

= \frac{1}{3}  \times 12 = 4

Also, the number of people in the line for dunking machine

=  2 x ( The number of people in line for Cakewalk)

=  2  x  4  = 8

Hence, the total number of people in line for the dunking machine is 8.

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Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif
algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64  

e) p_v =P(t_{118}  

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

\bar X_{O}=2.91 represent the mean for the Orange Coast

\bar X_{C}=2.96 represent the mean for the Coastline

s_{O}=0.05 represent the sample standard deviation for Orange Coast

s_{C}=0.03 represent the sample standard deviation for Coastline

n_{O}=60 sample size for Orange Coast

n_{C}=60 sample size for Coastline

\alpha=0.005 Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis:\mu_{O} \geq \mu_{C}  

Alternative hypothesis:\mu_{O} < \mu_{C}  

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=60+60-2=118  

Replacing the info we got:

t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64  

Part e

We can calculate the p value with this probability:

p_v =P(t_{118}  

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

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Marking brain list..
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Answer:

C

Step-by-step explanation:

You do 8 by 13 which is 104 now multiply that by 3 which is 312 then divide by 2 because it triangular

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Step-by-step explanation:

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