Need Help 2 & 3 with inequalities!!!

Every morning Lucas runs 1 7/10 miles. How far does he run in 5 days?

dybincka [34]

1 (5) = 5

7/10(5/1) = 35/10

Simplify.

35/10 = 3 5/10 = 3 1/2

5 + 3 1/2 = 8 1/2

Lucas ran 8 1/2 miles in 5 days.

7 0

3 years ago

Help pls!

Sliva [168]

Answer:5.83k

Step-by-step explanation:

annual means divide current by 6

3 0

3 years ago

This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex

noname [10]

$L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)$

$L_x=10+10\lambda x=0\implies1+\lambda x=0$

$L_y=10+10\lambda y=0\implies1+\lambda y=0$

$L_z=2+4\lambda z=0\implies1+2\lambda z=0$

$L_\lambda=5x^2+5y^2+2z^2-42=0$

$\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z$

$5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1$

$z^2=1\implies z=\pm1\implies x=y=\pm2$

There are two critical points, at which we have

$f\left(2,2,1\right)=42\text{ (a maximum value)}$

$f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}$

3 0

3 years ago

State and prove Bayes Theorem

Delvig [45]

Answer:

Bayes’ Theorem describes the probability of occurrence of an event related to any condition. It is also considered for the case of conditional probability.

For prove refer to the attachment.

Hope this helps you^_^

3 0

2 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

So

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

So

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0

3 years ago