Which absolute value equation represents the graph?

Write an arithmetic expression that calculates the average of 18 and 46. you may already know that: the average of two numbers i

Masteriza [31]

In this question , we have to write an arithmetic expression that calculates the average of 18 and 46.

TO find the average, we have to add the numbers and divide by 2.

So here we have to add 18 and 46 and divide by 2, that is

$d = \frac{18+46}{2} = \frac{64}{2}$

And that's the required algebraic expression .

8 0

4 years ago

HELPPPPPPPP 7 POINTS

Shalnov [3]

The correct answer is 5. This is because rate of change of a function or a line is its gradient. As we can see that the line is horizontal then the gradient must be equal to zero too.

On the other hand the equation is written in the form of y = mx + c where m is the gradient or slope and c is the y intercept so the gradient is equal to 5 if we match the places. So it is 5 times greater.

7 0

3 years ago

Geologists estimate the time since the most recent cooling of a mineral by counting the number of uranium fission tracks on the

Oliga [24]

Answer:

(a) The value of P (X = 7) is 0.1388.

(b) The value of P (X ≥ 3) is 0.9380.

(c) The value of P (2 < X < 7) is 0.5433.

(d) $\mu_{X}=6$

(e) $\sigma_{X}=2.45$

Step-by-step explanation:

Let X = number of uranium fission tracks on per cm² surface area of the mineral.

The average number of track per cm² surface area is, λ = 6.

The random variable X follows a Poisson distribution with parameter λ = 6.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, 3...$

(a)

Compute the value of P (X = 7) as follows:

$P(X=6)=\frac{e^{-6}(6)^{7}}{7!}=\frac{0.0025\times 279936}{5040}=0.1388$

Thus, the value of P (X = 7) is 0.1388.

(b)

Compute the value of P (X ≥ 3) as follows:

P (X ≥ 3) = 1 - P (X < 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2)

$=1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380$

Thus, the value of P (X ≥ 3) is 0.9380.

(c)

Compute the value of P (2 < X < 7) as follows:

P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

$=\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443$