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yuradex [85]
3 years ago
15

SOMEONE HELP PLEASE! 40 POINTS!!!! Answer these questions: 1. The relation {(-1, 4), (2, 7), (3, 7)} is a function. True False 2

. The function {(1, 4), (2, 5), (3, 6)} has the function rule y = 3 x + 1. True False 3. What is the value of the 11th term in the sequence -3, -6, -12, -24, ...? -6,144 -118,098 -3,072 -354,294 4. All of the following statements describe the relation shown except _____. The diagram represents a function. The inputs are {-1, 0, 3, 4}. The outputs are {-2, -1, 1}. This relation could also be represented as {(-1, 1), (0, -2), (3, 1), (4, -1)}. 5. What is the function rule represented by the following table? x y 0 3 1 -1 2 -5 y = -2 x + 1 y = -3 x y = -2 x - 1 y = -4 x + 3
Mathematics
1 answer:
DerKrebs [107]3 years ago
5 0

Answer:

1. True

2. False

3.-48

4. This relation could also be represented as {(-1, 1), (0, -2), (3, 1), (4, -1)}

5. can you please put the points in parentheses becasue I cannot tell which pair is which. If you rephrase it I will comment the anwer

Step-by-step explanation:

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1. What is the binomial expansion of (x + 3)5? ​
lara [203]

Answer:

5x + 15

Step-by-step explanation:

To expand the binomial, you need to multiply each term inside the parentheses by 5.

(x + 3)5

= (x * 5) + (3 * 5)

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2 years ago
Arandom sample of n1 =12 students majoring in accounting in a college of business has a mean grade-point average of 2.70 (where
antiseptic1488 [7]

Answer:

We accept the null hypothesis that the mean gpa's are equal, with the 5 percent level of significance.

Step-by-step explanation:

We have these following hypothesis:

Null

Equal means

So

\mu_{1} = \mu_{2}

Alternative

Different means

So

\mu_{1} \neq \mu_{2}

Our test statistic is:

\frac{\overline{Y_{1}} - \overline{Y_{2}}}{\sqrt{\frac{s_{1}^{2}}{N_{1}} + \frac{s_{2}^{2}}{N_{2}}}}

In which \overline{Y_{1}}, \overline{Y_{2}} are the sample means, N_{1}, N_{2} are the sample sizes and s_{1}, s_{2} are the standard deviations of the sample.

In this problem, we have that:

\overline{Y_{1}} = 2.7, s_{1} = 0.4, N_{1} = 12, \overline{Y_{2}} = 2.9, s_{2} = 0.3, N_{2} = 10

So

T = \frac{2.7 - 2.9}{\sqrt{\frac{0.4}^{2}{12} + \frac{0.3}^{2}{10}}} = -1.3383

What to do with the null hypothesis?

We will reject the null hypothesis, that is, that the means are equal, with a significante level of \alpha if

|T| > t_{1-\frac{\alpha}{2},v}

In which v is the number of degrees of freedom, given by

v = \frac{(\frac{s_{1}^{2}}{N_{1}} + \frac{s_{2}^{2}}{N_{2}})^{2}}{\frac{\frac{s_{1}^{2}}{N_{1}}}{N_{1}-1} + \frac{\frac{s_{2}^{2}}{N_{2}}}{N_{2} - 1}}

Applying the formula in this problem, we have that:

v = 20

So, applying t at the t-table at a level of 0.975, with 20 degrees of freedom, we find that

t = 2.086

We have that

|T| = 1.3383

Which is lesser than t.

So we accept the null hypothesis that the mean gpa's are equal, with the 5 percent level of significance.

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