The probabilities of the orphaned pets in six cities' animal shelters being different types of animals are given in the table. I

We need to specify the interval for numbers ranging from -5 to 4.

Moreover, we must exclude the values -1 and 4 from this range.

First, let’s note some crucial guidelines regarding interval notation.

It’s essential to identify which values are included and which ones are not in the specified range.

An open interval (or) indicates excluded numbers, while a closed interval [or] shows included numbers.

In this scenario, we notice three key numbers: -5, -1, and 4.

It is explicit that -1 and 4 are excluded, though -5 is not indicated as excluded, hence it remains included.

-1 lies between -5 and 4 on the number line.

Consequently, we can represent the interval notation from -5 to -1. (Note: -5 is included and -1 is not).

Thus, we write a closed interval " [ " on the left side of -5 and an open interval " ( " on the right side of -1. This gives us,

[-5, -1 ).

Next, we represent the interval from -1 to 4.

Since both endpoints are excluded, we will use open intervals ( ) on both sides of -1 and 4.

This results in the interval being

(-1,4).

We combine the two interval notations using the union symbol. So, the final interval representation is

[-5, -1 ) ∪ (-1,4).

Respuesta:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Explicación paso a paso:

La probabilidad (P) de encontrar la partícula está dada por:

 

    (1)

La solución de la integral de la ecuación (1) es:

 

(a) La probabilidad de encontrar la partícula entre x = 4.95 nm y 5.05 nm es:

   

(b) La probabilidad de encontrar la partícula entre x = 1.95 nm y 2.05 nm es:

 

(c) La probabilidad de encontrar la partícula entre x = 9.90 nm y 10.00 nm es:

   

(d) La probabilidad de encontrar la partícula en la mitad derecha de la caja, es decir, entre x = 0 nm y 50 nm es:

(e) La probabilidad de encontrar la partícula en el tercio central de la caja, es decir, entre x = 0 nm y 100/6 nm es:

Espero que te ayude.

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The equation of the line with the LARGEST slope that is tangent to both graphs is

(b) The equation of the second line tangent to both graphs is:

Solution:

- First, let's calculate the derivatives for the two functions provided:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Given that the derivatives of both functions are dependent on the x-value, we will pick a common point x_o for both f(x) and g(x). This point is ( x_o, g(x_o)). Thus,

                                g'(x_o) = -2*(x_o - 2)

- Next, we will determine the slope of a line that is tangent to both graphs at point (x_o, g(x_o) ) on g(x) and at ( x, f(x) ) on f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now we need to set the slope from our equation equal to the derivatives we calculated earlier for each function:

                                m = f'(x) = g'(x_o)

- We will work through the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now we can subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Rearranging gives us:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o,     x_o = 10x + 2    

- For x_o = 10x + 2,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Now solving the above quadratic equation:

                                 x = -0.0574, -0.387      

- The maximum slope occurs at x = -0.387, with the line’s equation being:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- The second tangent line is:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

To determine if there is evidence suggesting a change in average height, we can conduct a right-tailed test and formulate both null and alternative hypotheses.

H₀ (null hypothesis): μ = 162.5
H₁ (alternative hypothesis): μ > 162.5

With two samples to analyze, we can calculate the z-score using the formula provided below.



In this formula, Z symbolizes the z-score, Χ denotes the new sample mean, μ indicates the theoretical average, δ represents the standard deviation, and n signifies the sample size. Based on the gathered values, 




Assuming a significance level of α = 0.05. With a z-score of 2.77, we can reference the z-table to ascertain the p-value. This yields P(Z > 2.77) =.0028. Since our p-value is below α, we reject the null hypothesis, indicating that the average height of female freshman students has indeed shifted.