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s2008m [1.1K]
3 years ago
11

2. What is the difference between 4 and -5?

Mathematics
1 answer:
adell [148]3 years ago
8 0

Answer: 9

Step-by-step explanation: 4 - (-5) = 9

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A company rounds its losses to the nearest dollar. The error on each loss is independently and uniformly distributed on [–0.5, 0
lesya [120]

Answer:

the 95th percentile for the sum of the rounding errors is 21.236

Step-by-step explanation:

Let consider X to be the rounding errors

Then; X \sim U (a,b)

where;

a = -0.5 and b = 0.5

Also;

Since The error on each loss is independently and uniformly distributed

Then;

\sum X _1 \sim N ( n \mu , n \sigma^2)

where;

n = 2000

Mean \mu = \dfrac{a+b}{2}

\mu = \dfrac{-0.5+0.5}{2}

\mu =0

\sigma^2 = \dfrac{(b-a)^2}{12}

\sigma^2 = \dfrac{(0.5-(-0.5))^2}{12}

\sigma^2 = \dfrac{(0.5+0.5)^2}{12}

\sigma^2 = \dfrac{(1.0)^2}{12}

\sigma^2 = \dfrac{1}{12}

Recall:

\sum X _1 \sim N ( n \mu , n \sigma^2)

n\mu = 2000 \times 0 = 0

n \sigma^2 = 2000 \times \dfrac{1}{12} =  \dfrac{2000}{12}

For 95th percentile or below

P(\overline X <  95}) = P(\dfrac{\overline X - \mu }{\sqrt{{n \sigma^2}}}< \dfrac{P_{95}- 0 } {\sqrt{\dfrac{2000}{12}}}) =0.95

P(Z< \dfrac{P_{95} } {\sqrt{\dfrac{2000}{12}}}) = 0.95

P(Z< \dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}}) = 0.95

\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1- 0.95

\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} = 0.05

From Normal table; Z >   1.645 = 0.05

\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1.645

{P_{95}\sqrt{12} } = 1.645 \times {\sqrt{{2000}}}

{P_{95} = \dfrac{1.645 \times {\sqrt{{2000}}} }{\sqrt{12} } }

\mathbf{P_{95} = 21.236}

the 95th percentile for the sum of the rounding errors is 21.236

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3 years ago
If there are 16 boys and 32 girls in a room, fill out all of the possible ratios of boys to girls that could be made. They alrea
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All of possible ratios of boys to girls that could be made is 8:16; 4:8; 2:4; 1:2
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What is 3x^3 - 11x^2 - 26x + 30 divided by x-5?
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Answer:

Most likely the answer is

3x^2+4x-6

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2000mL - 280mL = 1720mL or 1.72L
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