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Igoryamba
3 years ago
7

Use the long division method to find the result when 6x3 – 22x2 + 19x - 8 is

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

5964672

Step-by-step explanation:

if this isn't the answer then I don't know what it is I'm not 100% sure

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Does the given matrix, A, have an inverse? If it does, what is A-1?
ololo11 [35]

Answer:

yes he does

Step-by-step explanation:

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3 years ago
Someone please help!
Vlad1618 [11]
Multiply the top equation by 6 and the bottom equation by 7. You will cancel x first by doing this and solve for y. You should get y=-1.
Substitute in y for either original equation.
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3 years ago
Why is the formula for the area of a triangle one half the area for a rectangle
Naya [18.7K]
The area of a triangle is one half the area for a rectangle because the triangle is half of a rectangle with equal base and height
8 0
3 years ago
Answer the questions please
Drupady [299]

Answer:

4.  988 people

5. a. 7 quarters

    b. 6 quarters 2 dimes 1 nickel

    c. 5 quarters 5 dimes

    d. 4 quarters 7 dimes 1 nickel

1.  $65.65, $5.05

2. $7.70, $15.40

3. a. 7 quarters

    b. 6 quarters 2 dimes 1 nickel

    c. 5 quarters 5 dimes

    d. 4 quarters 7 dimes 1 nickel

4. 31 bags

5. $30, $4.50

Step-by-step explanation:

4. take value times percent.

5. think coins

1. find difference then divide by number of days (13)

2. multiply by 1/3 then find difference.

3. think coins

4. find area then divide by 200

5. just break the problem down and go quarter by quarter.

4 0
4 years ago
Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0
3 years ago
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