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WINSTONCH [101]
3 years ago
5

WILL GIVE BRAINLIEST :)

Mathematics
1 answer:
Nonamiya [84]3 years ago
7 0
Step One
Develop Pascal's Triangle
                              1
                           1     1
                         1    2    1
                        1  3    3    1
                      1  4   6    4    1
                     1  5  10  10  5   1    <<<< How the 5th power expands.

Step Two
Confirm that nCr gives the same thing as the triangle gives.

The third term is outlined as 10 a^3 b^2 The powers must add to 5.
The 10 can be found this way

nCr Where n is the row number and r is the term you want so
n = 5
r = 3
nCr = n! / [(n - r)! r!]
5C3 = 5! / ((3!) (2!) )
5C3 = 5 * 4 / 2
5C3 = 10 Just as the triangle predicted.

Step Three
Find a^3 b^2
a = 3x
b = - 5y

a^3 = (3x)^3
a^3 = 27x^3
b^2 = (-5y)^2
b^2 = 25 y^2

Step Three
Find the Answer
Third term = 10*27x^3*25y^2
Third term = 6750 x^3y^2

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2 years ago
Textbooks are exempted from PST in Ontario. If the total cost of a book
Andrei [34K]

Answer:

4.24

Step-by-step explanation:

Total price = price of the book + 5% of the price of the book

T = total price   P = price of the book G = GST

T = P + G

and

G = 5% of P = 0.05P

therefore

T = P + 5%P = 1P + 0,05P = 1.05P

P = T/1.05 = 89/1.05 ≅ 84.76 (it is 84.7619047619)

G = 89 - 84.76 = 4.24

8 0
3 years ago
Analyze the results of the correlation using the given residual plot.
Xelga [282]

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Step-by-step explanation:

7 0
2 years ago
A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In
lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

                        P.Q. = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of males having blood disorder= \frac{250}{1000} = 0.25

\hat p_2 = sample proportion of females having blood disorder = \frac{275}{1000} = 0.275

n_1 = sample of males = 1000

n_2 = sample of females = 1000

p_1 = population proportion of males having blood disorder

p_2 = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

= [ (0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }, (0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} } ]

 = [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0
3 years ago
Family bought 12 oranges from the market. However, one-fourth of these oranges were rotten. How many oranges were not rotten?
sashaice [31]

Answer:

1/4 of them were rotten

rotten ones = 1/4 × 12 = 3

ones that weren't rotten = 12 - 3 = 9

4 0
2 years ago
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