What is f(–3) for the function f(a) = –2a2

aalyn [17]

1 pound = 16 ounces
multiply the amount of pounds you have by 16 (that's how many ounces is in a pound) your answer should be whatever you get
example: you have 13 pounds, just multiply it by 16, the answer is 208.
so there are 208 ounces in 13 pounds.
or you could go google and ask them.
hope this helps.

8 0

3 years ago

I need help with 38,39 and 40 please!

kodGreya [7K]

Answer:

It's 37. B 38. B 39. C 40. A

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ. $e^{-\lambda.x}$

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - $e^{-\lambda.x}$

(a) Probability of distance at most 100m, with λ = 0.0143:

F(100) = 1 - $e^{-0.0143.100}$

F(100) = 0.76

Probability of distance at most 200:

F(200) = 1 - $e^{-0.0143.200}$

F(200) = 0.94

Probability of distance between 100 and 200:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = $\frac{1}{\lambda}$

E(X) = $\frac{1}{0.0143}$

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = $\sqrt{\frac{1}{\lambda^{2}} }$

σ = $\sqrt{\frac{1}{0.0143^{2}} }$

σ = 69.93

Distance exceeds the mean distance by more than 2σ:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - $e^{-0.0143*209.79}$ )

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

$\int\limits^m_0 f({x}) \, dx$ = 0.5

$\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx$ = 0.5

$\lambda.\frac{e^{-\lambda.x}}{-\lambda}$ = $-e^{-\lambda.x} + e^{0}$

$1 - e^{-\lambda.m}$ = 0.5

$-e^{-\lambda.m}$ = - 0.5

ln( $e^{-0.0143.m}$ ) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0

4 years ago

The area of sector A08 is 48x and m<AOB = 270º. Find the radius of circle O.

Softa [21]

Answer:

Step-by-step explanation:

The first one gives us everything we need except the radius, which is easy enough to solve for if you're careful with your algebra. The area of a sector of a circle is given as:

$A_s=\frac{\theta}{360}*\pi r^2$ where θ is the measure of the central angle of the circle. For us, that fills in as follows:

$48\pi=\frac{270}{360}*\pi r^2$ and manipulate it as follows:

$r^2=\frac{(360)(48\pi)}{270\pi}$ the π's cancel out, leaving us with simple multiplication and division to get

r = 8. Now for the next one, which is a bit more involved.

In order to find the area of the shaded part, we need to find the area of the right triangle there and subtract it from the area of the sector of the circle. First the area of the sector, which is given as:

$A_s=\frac{\theta}{360}*\pi r^2$ where θ again is the measure of the central angle of the circle, 90°:

$A_s=\frac{90}{360}(3.1415)(10)^2$ which simplifies a bit to

$A_s=\frac{1}{4}(3.1415)(100)$ , giving us an area of

$A_s=78.5375m^2$ . Now onto the area of the triangle.

Since this triangle is inscribed in the circle and the circle's radius is 10, tha also gives us both the height and the base measures of the triangle. The area then is:

$A_t=\frac{1}{2}(10)(10)$ which is

$A_t=50m^2$

Subtract that from the area of the sector to get that the shaded area is 28.5 square meters, choice A.

4 0

3 years ago

Need helpppppppppp! Show work plz

Natasha_Volkova [10]

Answer:

9/169

Step-by-step explanation:

JQK of 4 suits = 12 cards

12/52x12/52=

3/13x3/13=

9/169

5 0

3 years ago

What is f(–3) for the function f(a) = –2a2 – 5a + 4?