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2 years ago

write the equation of the line that is parallel to the line represented by 5 x + 2 y = 6 and passes through the point (1, 5.5).

Mathematics

1 answer:

Oliga [24]2 years ago

3 0

Answer:

$y=-2.5x+8$

Step-by-step explanation:

Parallel lines have the same slope but different y-intercepts. So, first, transform the given equation into the slope-intercept form, $y=mx+b$ .

$5x+2y=6$

$2y=-5x+6$

$y=-\frac{5}{2} x+3$

Then, determine the equation of a line that has a slope of $-\frac{5}{2}$ and passes through (1, 5.5). Substitute all the known values and solve for b.

$y=mx+b$

$5.5 = -\frac{5}{2} *1+b$

$5.5=-2.5+b$

$b=8$

Therefore, the answer is $y=-2.5x+8$ .

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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

8 0

3 years ago

Which linear inequality represents the solution set graphed?

postnew [5]

Answer:

Correct option is C

Step-by-step explanation:

First, find the equation of the boundary line. This line passes through the points (0,-3) and (-2,1). Then it has equation

$\dfrac{x-0}{-2-0}=\dfrac{y+3}{1+3},\\ \\4x=-2(y+3),\\ \\y=-2x-3.$

In the attached diagram this boundary line is solid, then the sign of the inequality should be with "or equal to" notion (≤ or ≥). Thus, options A and B are false.

The line divides the coordinate plane into two parts and you have to determine which part to choose. The origin does not lie in the shaded region, then its coordinates cannot satisfy the inequality. Check options C and D.

$0\le -2\cdot 0-3\ (0\le -3)$ - origin does not satisfy (option C is correct);