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3 years ago

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Jayden scored more than 2/3 the number of points Kenneth scored. If Jayden scored 24 points, which inequality solution represent

s , the number of points Kenneth could have scored?

1 answer:

natka813 [3]3 years ago

4 0

Answer:The answer would be A. 1 1/5

48/40 = 1 8/40

1 8/40 Simplify to 1 1/5

Hope this helps!

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Sally took out a $50,000 life insurance policy. The $50,000 amount of coverage is called the:

olga55 [171]

Answer:

The $50,000 amount of coverage is called the Death benefit or face value.

Step-by-step explanation:

Consider the provided information.

Sally took out a $50,000 life insurance policy.

The face value, or death benefit is the amount of money a life insurance policy would pay upon the insured person's death to the beneficiary.

Hence, the $50,000 amount of coverage is called the Death benefit or face value.

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3 years ago

Complete the Magic Square.

quester [9]

Answer:

+5

Step-by-step explanation:

So you can see the 5 pattern, so you add 9+5= 14 which goes in the middle square and 14+5=18 in the left square

Now between 18 and 11 it is 7 so you subtract 7 which equals 4 then - 5= -1 then -5= -6

5 0

3 years ago

Read 2 more answers

Cheyenne's bi-weekly gross pay is $529.81. She sees that $18.54 was deducted for Medicare tax. What percent of Cheyenne's gross

Dafna1 [17]

18.54/529.81= 0.0349
0.0349*100=3.49%= 3.5%

6 0

3 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =

Taya2010 [7]

Answer:

$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

<em>Maximum value of f=2.41</em>

Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

$\bigtriangledown f=\lambda \bigtriangledown g$

for some scalar $\lambda$ called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

$\bigtriangledown f=\lambda \bigtriangledown g+\mu \bigtriangledown h$

The gradient of f is

$\bigtriangledown f=$

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in $x,y,z,\lambda,\mu$ .

We have

$f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0$

Let's compute the partial derivatives

$f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0$

The Lagrange condition leads to

$1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)$

Operating and simplifying

$1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11$

Replacing the value of $\lambda$ in the two first equations, we get

$1=11+2x\mu\\2=-11 +2y\mu$

From the first equation

$\displaystyle 2\mu=\frac{-10}{x}$

Replacing into the second

$\displaystyle 13=y\frac{-10}{x}$

Or, equivalently

$13x=-10y$

Squaring

$169x^2=100y^2$

To solve, we use the restriction h

$x^2 + y^2 = 1$

Multiplying by 100

$100x^2 + 100y^2 = 100$

Replacing the above condition

$100x^2 + 169x^2 = 100$

Solving for x

$\displaystyle x=\pm \frac{10}{\sqrt{269}}$

We compute the values of y by solving

$13x=-10y$

$\displaystyle y=-\frac{13x}{10}$

For

$\displaystyle x= \frac{10}{\sqrt{269}}$

$\displaystyle y= -\frac{13}{\sqrt{269}}$

And for

$\displaystyle x= -\frac{10}{\sqrt{269}}$

$\displaystyle y= \frac{13}{\sqrt{269}}$

Finally, we get z using the other restriction

$x - y + z = 1$

Or:

$z = 1-x+y$

The first solution yields to

$\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

And the second solution gives us

$\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Complete first solution:

$\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

5 0

3 years ago

The following table shows the probability distribution for a discrete random variable.

mezya [45]

Answer: C. 30.47

Step-by-step explanation:

The mean of discrete random variable i.e. the expected value of X is given by :-

$E(X)=\sum^{i=n}_{i=1} X_iP(X_i)$

Now by using the given table, the expected value of X is given by :-

$\\\\\Rightarrow\ E(X)=23\cdot (0.16)+25\cdot(0.09)+26\cdot(0.18)+31\cdot(0.12)+34\cdot(0.24)+38\cdot(0.21)\\\\\Rightarrow\ E(X)=30.47$

Hence, the mean of discrete random variable= 30.47

6 0

3 years ago