All categories

3 years ago

Can some also explain it to me how to identify if it’s continuous exponential

Mathematics

1 answer:

sweet-ann [11.9K]3 years ago

8 0

B is the correct answer

hope this helps

You might be interested in

Which expression is not a polynomial? 11p-7q 8xy^4-5x^3y 2-x -4/y

kkurt [141]

A polynomial is an expression that has two or more than terms. These terms are separated by an operation. Only -4/y is not a polynomial. It is a monomial. Monomial is not a polynomial.

3 0

3 years ago

The accompanying data represent the daily (for example, Monday to Tuesday) movement of Johnson & Johnson (JNJ) stock for

egoroff_w [7]

Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

The lower bound is of 0.484.
The upper bound is of 0.7292.

The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

Supposing that it increases on 37 out of 61 days:

$n = 61, \pi = \frac{37}{61} = 0.6066$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 - 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.484$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 + 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.7292$

The 95% confidence interval for the proportion of days JMJ stock increases is (0.484, 0.7292), in which 0.484 is the lower bound and 0.7292 is the upper bound.

The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

A similar problem is given at brainly.com/question/16807970

4 0

3 years ago

Which best describes the figure that is represented by the following coordinates?

kolezko [41]

Answer: its square.

Step-by-step explanation:

7 0

3 years ago

What numbers make this expression undefined x^2+3/x^2-11x-12

marusya05 [52]

$\frac{ {x}^{2} + 3}{ {x}^{2} - 11x - 12}$

Because which numbers will make the expessions undefined. We will focus on the denominator only.

Factor the terms.

$(x - 12)(x + 1)$

Now which 2 numbers that will make these terms become 0?

12 and - 1 of course.

So these 2 numbers that will make the expressions undefined are 12 and - 1

5 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).