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liq [111]
1 year ago
14

Determine the domain and the range of the relation. The domain for this relation is (0, 1, 3). The range for this relation is (-

2, 1, 2)
Mathematics
1 answer:
dlinn [17]1 year ago
6 0

The domain for this relation is {1,3} and the range for this relation is {2}.

According to the question, the given domain and range of the relation is as follows:

The domain for this relation is : {0, 1, 3}

The range for this relation is : {-2, 1, 2}

As per definition, the domain is the set of all possible values which a function take as an input.

Therefore, the domain for this relation is : {1, 3}

And the range for this relation is: {2}

Hence, the domain for this relation is {1,3} and the range for this relation is {2}.

What is domain and range of a function?

The domain is the set of all the possible values which a function can take as an input. And the range is also the set of paired values which represent the output of a function.

To learn more about the domain and range of a function from the given link:

<u>brainly.com/question/1942755</u>

#SPJ9

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1, 2, 3, 4, 5, 6, and 12

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Help me with this please
ioda
1) not function because 6 number repeat in domain
2) Function
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6 0
3 years ago
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

5 0
3 years ago
You borrowed $59,000 for 2 years at 11% which was compounded annually. What
gayaneshka [121]

Answer:

$72693.9

Step-by-step explanation:

To get this answer you need to use the compound interest formula, which will be A=P(1+r/n)^n(t). P=59,000 r=11%=0.11 n=1 (annually) t=2 years. From there you should be able to figure the rest out and get the answer. Hope this helps!

6 0
2 years ago
Please help really need to pass
VashaNatasha [74]
First add 5 to right side
\frac{x}{7} = -4+5 = 1
Next , to cancel the 7, you need to multiply by 7 on both sides
x = 1*7 = 7
8 0
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