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navik [9.2K]
3 years ago
15

PLEASE HELP!! WILL MARK BRAINLEST!!

Mathematics
2 answers:
posledela3 years ago
7 0
$698.09
you would multiply 638.99 multiplied by 9.25% then take that answer and add it to the original price
ra1l [238]3 years ago
6 0

Answer:

I'm sorry I tried to solve this, but my answer did not seem correct. You do not need to name me brainliest, but here is a link that might help you https://www.mathgoodies.com/lessons/percent/sales_tax

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Using the given points, you can write the equation as ...
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What is p^m ÷ p^n equal to?<br> pm − n <br> pm + n <br> pm • n <br> pm ÷ n
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EXPONENTIALLY \: \: \: REDUCED \\ \\ \\ \\Given \: \: Expression \: \: - \\ \\ \\ \: \: \: \: \: \: \: \: \: \: {p}^{m} \: \: \div \: \: {p}^{n} \\ \\ \\ \: \: \: = \: \: \frac{ {p}^{m} }{ {p}^{n} } \\ \\ \\ \\ \: \: \: = \: \: {p}^{m - n} \: \: \: \: \: \: \: \: \: \: Ans.
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Answer:

There is no correct answer

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4x+20

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4x+20=2x+6

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5 0
3 years ago
Solve for u . 3u+11=41
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5 0
3 years ago
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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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