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2 years ago

How many three-letter "words" can Bob make using just these letters?

Mathematics

1 answer:

alekssr [168]2 years ago

4 0

Answer:

two

Step-by-step explanation:

the two "words" Bob can make using the three letters t, o, and a are;

oat and tao

hope this helps

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The

adoni [48]

The critical value is the z-score right at the edge of the rejection region option (A) is correct.

<h3>What is a normal distribution?</h3>

It's the probability curve of a continuous distribution that's most likely symmetric around the mean. On the Z curve, at Z=0, the chance is 50-50. A bell-shaped curve is another name for it.

It is given that:

The z-score is right at the edge of the rejection region.

As we know, the rejection zone is the area in which we have sufficient data to reject the null hypothesis if our test statistic falls within it.

Thus, the critical value is the z-score right at the edge of the rejection region option (A) is correct.

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1 year ago

guse lagrange multipliers to find the maximum or minimum values of the function subject to the given constraint. (if an answer d

Novosadov [1.4K]

Therefore the maximum value of function f(x,y,z)= $x^{2} y^{2} z^{2}$ =1/27

And the minimum value is 0

<h3>What is function?</h3>

function, in mathematics, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable) (the dependent variable) (the dependent variable) (the dependent variable). Mathematics uses functions frequently, and functions are essential for specifying physical relationships in the sciences.

Here,

The function is given as:

f(x,y,z)= $x^{2} y^{2} z^{2}$

$x^{2} +y^{2}+ z^{2}$ =1

=> $x^{2} +y^{2}+ z^{2}$ -1=0

Using Lagrange multiplies, we have:

L(x,y,z,λ)=f(x,y,z) +λ(0)

Substitute f(x,y,z)= $x^{2} y^{2} z^{2}$ and $x^{2} +y^{2}+ z^{2}$ -1=0

Differentiate

L(x)= $2xy^{2} z^{2}$ +2λx

L(y)= $2yx^{2} z^{2}$ +2λy

L(z)= $2zx^{2} y^{2}$ +2λz

L(λ)= $x^{2} +y^{2}+ z^{2}$ -1

Equating to 0

$2xy^{2} z^{2}$ +2λx =0

$2yx^{2} z^{2}$ +2λy = 0

$2zx^{2} y^{2}$ +2λz = 0

$x^{2} +y^{2}+ z^{2}$ -1 = 0

Factorize the above expressions

$2xy^{2} z^{2}$ +2λx =0

2x( $y^{2} z^{2}$ +λ)=0

2x=0 and ( $y^{2} z^{2}$ +λ)=0

x=0 and $y^{2} z^{2}$ = -λ

$2yx^{2} z^{2}$ +2λy = 0

2y( $x^{2} z^{2}$ +λ)=0

2y=0 and ( $x^{2} z^{2}$ +λ)=0

y=0 and $x^{2} z^{2}$ = -λ

$2zx^{2} y^{2}$ +2λz = 0

2z( $y^{2} x^{2}$ +λ)=0

2z=0 and ( $x^{2} y^{2}$ +λ)=0

z=0 and $x^{2} y^{2}$ = -λ

So we have ,

x=0 and $y^{2} z^{2}$ = -λ

y=0 and $x^{2} z^{2}$ = -λ

z=0 and $x^{2} y^{2}$ = -λ

The above expression becomes

x=y=z=0

This means that,

$x^{2} +y^{2}+ z^{2}$ =1

$x^{2} +x^{2}+ x^{2} =1 \\3x^{2 } =1$

x= ±1/ $\sqrt{3}$

So,

y= ±1/ $\sqrt{3}$

z= ±1/ $\sqrt{3}$

The critic points are

x=y=z=±1/ $\sqrt{3}$

x=y=z=0

Therefore the maximum value of function f(x,y,z)= $x^{2} y^{2} z^{2}$ =1/27

And the minimum value is 0

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1 year ago

3+12x5/(5+15) What’s the answer

r-ruslan [8.4K]

$\bf 3+12\times \cfrac{5}{(5+15)}\implies 3+12\times \cfrac{5}{\underset{\uparrow }{20}}\implies 3+12\times \cfrac{1}{4}\implies 3+\cfrac{\stackrel{\downarrow }{12\times 1}}{4} \\\\\\ 3+\cfrac{12}{4}\implies 3+\stackrel{\downarrow }{3}\implies 9$