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Tpy6a [65]
3 years ago
15

1. find the degree of the monomial 6p^3q^2

Mathematics
2 answers:
allsm [11]3 years ago
4 0

Answer:

Part 1: 5

Part 2: 13t^2+17

Part 3:12n^3+15n^2

Part 4:9(2s-7)

Part 5: 25x^6-4

Part 6: (x^2+12x+35)

Part 7: 4z^2-12z+9

Step-by-step explanation:

Part 1: we have to find the degree of monomial 6p^3q^2

The degree of a polynomial is the highest degree of its monomials (individual terms) with non-zero coefficients.

The degree of monomial= 3+2=5

Option c is correct.

Part 2:  simplify (7t^2+9) + (6t^2+8)

(7t^2+9) + (6t^2+8)

(7t^2+6t^2)+(9+8)

13t^2+17

Option b is correct.

Part 3: simplify 3n(4n^2+5n)

3n(4n^2+5n)

By distributive property

12n^3+15n^2

Option a is correct.

Part 4: . factor 18s-63

18s-63

Taking 9 common from both terms

9(2s-7)

Option d is correct.

Part 5: simpler form of (5x^3+2)(5x^3-2)

(a-b)(a+b)=a^2-b^2

(5x^3+2)(5x^3-2)=(5x^3)^2-2^2=25x^6-4

Option d is correct.

Part 6:  simplify (x+7)(x+5)

(x+7)(x+5)=(x^2+5x+7x+35)=(x^2+12x+35)

Option a is correct.

Part 7: simplify (2z-3)^2

(a-b)^2=a^2+b^2-2ab

(2z-3)^2=(4z^2+9-2(2z)(3)=4z^2-12z+9

Option d is correct.

natali 33 [55]3 years ago
3 0

Theses are the answers for connections.

1.C

2.B

3.A

4.D

5.D

6.A

7.D

Hope this helps! Have a great day!

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Alternative hypothesis:\mu_{Visa} \neq \mu_{Mastercard}

t=\frac{66970-59060}{\sqrt{\frac{9500^2}{11}+\frac{10000^2}{17}}}}=2.108  

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Data given and notation

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Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:\mu_{Visa}=\mu_{Mastercard}

Alternative hypothesis:\mu_{Visa} \neq \mu_{Mastercard}

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

z=\frac{\bar X_{Visa}-\bar X_{Masterdcard}}{\sqrt{\frac{s^2_{Visa}}{n_{Visa}}+\frac{s^2_{Mastercard}}{n_{Mastercard}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

t=\frac{66970-59060}{\sqrt{\frac{9500^2}{11}+\frac{10000^2}{17}}}}=2.108  

What is the p-value for this hypothesis test?

First we need to calculate the degrees of freedom given by:

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Since is a bilateral test the p value would be:

p_v =2*P(t_{26}>2.108)=0.0448

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.1 we see that p_v so we can conclude that we can reject the null hypothesis, and a would be a significant difference between the  in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold at 10% of significance .

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