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3 years ago

8

Explain why -0.33 is greater than -0.33. (This is not part of the question but there is a line above the second -0.33, like the

repeating line above)

1 answer:

Drupady [299]3 years ago

6 0

The line means it is -.3333333333333333 and so on , that's why it is greater than -.33

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What is the equation of a line passing between the points below? (-3,1) (5,-7)

kirill [66]

Answer:

Ax + Ay + 2A = 0 for any nonzero A, for example (A=1): x + y + 2 = 0

Step-by-step explanation:

The equation of a line is

Ax + By + C = 0

so we know that:

A*-3 + B*1 + C = 0

A*5 + B*-7 + C = 0

Let's subtract one from the other:

A*(-3 - 5) + B*(1 + -7) = 0

A*-8 + B*8 = 0

B*8 = A*8

B = A

Let's input B = A into the first two equations

A*-3 + A*1 + C = A*-2 + C = 0

A*5 + A*-7 + C = A*-2 + C = 0

checks out

C = 2A

So for any nonzero A the equation of

Ax + Ay + 2A = 0 produces a line passing between the points. Example would be

x + y + 2 = 0

4 0

3 years ago

PLEASE HELP ME !!!!!!

Ad libitum [116K]

Step-by-step explanation:

In this question as two straight lines intersect each other

we use vertically opposite angle axiom

so 74 degree = 2x.

then x= 74/2= 37degree

6 0

3 years ago

Read 2 more answers

How to solve -x^2-2x+1

Wittaler [7]

Answer:

Common factor

Factor by grouping

Factor by grouping

−

2

−

2

+

1

-x^{2}-2x+1

−x2−2x+1

−

1

(

2

+

2

−

1

)

{\color{#c92786}{-1(x^{2}+2x-1)}}

−1(x2+2x−1)

Solution−

-1(x²+ 2x-1 )

5 0

3 years ago

The total claim amount for a health insurance policy follows a distribution with density function 1 ( /1000) ( ) 1000 x fx e− =

gizmo_the_mogwai [7]

Answer:

the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

Step-by-step explanation:

The probability of the density function of the total claim amount for the health insurance policy is given as :

$f_x(x) = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0$

Thus, the expected total claim amount $\mu$ = 1000

The variance of the total claim amount $\sigma ^2 = 1000^2$

However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100

To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :

P(X > 1100 n )

where n = numbers of premium sold

$P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})$

$P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})$

$P(X>1100n) = P(Z> \dfrac{10*100}{1000})$

$P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345$

$\mathbf{P(X>1100n) = 0.158655}$

Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is $\mathbf{P(X>1100n) = 0.158655}$

4 0

3 years ago

the diagram shows the dimencions. 40yd 20yd 5yd 10yd 30yd. of a new parking lot. what is the area of the parking lot

Furkat [3]

1'200'000 yards because all you have to do is multiply the numbers.

3 0

3 years ago