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2 years ago

In how many ways can you put seven marbles in different colors into two jars? Note that the jars may be empty.

Mathematics

1 answer:

ahrayia [7]2 years ago

3 0

Answer:

128

Step-by-step explanation:

You have two jars and seven marbles.

You do $2^{7}$ .

That is 128

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When a certain basketball player takes his first shot in a game he succeeds with probability 1/2. If he misses his first shot, h

Gnom [1K]

Answer:

we are given

basketball player Chauncey Billups of the Detroit Pistons makes free throw shots 88% of the time

so, probability of making shot is

=88%

so, p=0.88

To find the probability of missing first shot and making the second shot

so, we can use formula

probability = p(1-p)

now, we can plug values

we get

So, the probability that he misses his first shot and makes the second is 0.1056........Answer

Step-by-step explanation:

3 0

2 years ago

2. The seniors at our high school decided to play a prank on the principal by completely filling his office with

melisa1 [442]

Answer:

idk

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?

irakobra [83]

First, note that $m\angle A+m\angle C=90^{\circ}.$ Then

$m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.$

Consider all options:

$\tan A=\dfrac{\sin A}{\sin C}$

By the definition,

$\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.$

Now

$\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.$

Option A is true.

$\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.$

Then

$\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.$

Option B is false.

$\sin C = \dfrac{\cos A}{\tan C}.$

By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

Now

$\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.$

Option C is false.

$\cos A=\tan C.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

As you can see $\cos A\neq \tan C$ and option D is not true.

$\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.$

By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.$

Then

$\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.$

This option is false.

8 0

2 years ago

Read 2 more answers

To make f continuous at x=2 , f(2) should be defined as what value? Justify your answer.

yawa3891 [41]

Answer:

To make f continuous at $x=2$ , $f(2)$ should be defined $x = 2.$

Step-by-step explanation:

A function, let say $f(x)$ , is defined at $x = c$ is continuous at $x = c$

If the limit of $f(x)$ as $x$ approaches $c$ is equal to the value of $f(x)$ at $x = c$ .

Mathematically it is written as:

$\lim _{x\to c}\:f\left(x\right)=f\left(c\right)$

then

$f(x)$ is continuous at $x = c$ .

So from the above definition, we conclude that:

To make f continuous at $x=2$ , $f(2)$ should be defined $x = 2.$

i.e.

$\lim _{x\to 2}\:f\left(x\right)=f\left(2\right)$

then

$f(x)$ is continuous at $x = 2$ .

8 0

1 year ago

If a(x) = 3x + 1 and b (x) = StartRoot x minus 4 EndRoot, what is the domain of (b circle a) (x)?

kiruha [24]

Answer:

$[1,\infty)$

Step-by-step explanation:

$b(x)=\sqrt{x-4}$

$a(x)=3x+1$

Since we want to know the domain of $(b \circ a)(x)$ , let's first consider the domain of the inside function, that is, that of $a(x)=3x+1$ . Every polynomial function has domain all real numbers.

So we can plug anything for function $a$ and get a number back.

Now the other function is going to be worrisome because it has a square root. You cannot take square root of negative numbers if you are only considering real numbers which that is the case with most texts.

Let's find $(b \circ a)(x)$ and simplify now.

$(b \circ a)(x)$

$b(a(x))$