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3 years ago

Please help me please?! ❤️❤️

Mathematics

1 answer:

Serhud [2]3 years ago

5 0

Answer:

(4,3)

Step-by-step explanation:

Point V lies in the Ist quadrant . and the co-ordinates are (4,3)

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What is the complement of 13 degrees

AlekseyPX

Since complementary angles add up to 90 degrees, we simple take $90=13+c$ and simplify it to $c=77$ °

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3 years ago

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Clark's weight is increasing these days if it's weight increases by 2 kg mg every 4 months how many kilograms of weight will he

ycow [4]

Answer:

8kg

Step-by-step explanation:

4m=2kg

16m=?

=16m×2

=32

=32÷4

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Hope this will help ya!

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3 years ago

Simplify. Show your work. 5 1/3 +(-3 9/18)

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3 years ago

Write this equation c + 25/ 100c as a coefficient fraction PLZ HELP !!! Really Apreciate

kogti [31]

Answer: 5/4c

Step-by-step explanation:

Hi, to answer this question we simply have to solve the equation given:

Coefficients are the numbers next to the variable; in this case, the variable is c.

c + 25/ 100c

1/1 c +25/100c

100/100c +25/100 c

125/100 c

5/4c

Feel free to ask for more if needed or if you did not understand something.

5 0

3 years ago

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

8 0

3 years ago