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bulgar [2K]
3 years ago
7

The sum of three numbers is 86 the second number is 9 less than the first the third number is 3 times the first what are the num

bers
Mathematics
1 answer:
Kobotan [32]3 years ago
7 0

Answer:

10, 19 and 57

Step-by-step explanation:

let the first number be a, the second be b and the third be c

a+b+c=86

the second number is 9 less than the first

a=b+9------i

b=a-9

the third number is 3 times the first

c=3a------ii

a=c/3

therefore

b+9=c/3

c=3(b+9)

b+9+b+3(b+9)=86

b+9+b+3b+27=86

collecting like terms

5b=86-(27+9)

5b=86-36

5b=50

therefore

b=10

recall i to solve for a

a=b+9

a=10+9

a=19

recall ii to solve for c

c=3a

c=3*19

c=57

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Step-by-step explanation:

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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  

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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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