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3 years ago

Math help pleaseeeeeeeeeeeeeeeeeeeeeeee

Mathematics

1 answer:

sleet_krkn [62]3 years ago

3 0

Answer:

b) For the given explicit definition, t(4) = 8

Step-by-step explanation:

Here, the given expression is:

$t_n = n^2 - 2n$

Now, we need to evaluate t(4).

Putting the value of n = 4, in the given expression, we get

$t = 4 \implies t_{(4)} = (4)^2 - 2(4)\\= 16 - 8 = 8\\\implies t_4 = 8$

Hence, for the given explicit definition, t(4) = 8

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kherson [118]

Answer:

The answer would be A/B=H

Step-by-step explanation:

To get rid of the B that is connected to the H, you would do the opposite of what they used. Meaning since the two were multiplied you would divide them. So, you would take your b and divide it on both sides of the equal sign. B and B cancels each other out so you are left with A/B which equals H.

7 0

2 years ago

There are 4 green marbles and 2 red marbles in the jar. You just randomly draw one by one without replacement and stop when you

zalisa [80]

Answer:

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3) = 2/15

P(4) = 1/15

The expected value for X is:

EV = 1.33...

Step-by-step explanation:

We have a total of 6 marbles in the jar.

The probability of getting a red marble in the first try (X = 0) is equal to the quotient between the number of red marbles and the total number of marbles, this is:

P(0) = 2/6 = 1/3

The probability of drawing one green marble (X = 1)

is:

First, you draw a green marble with a probability of 4/6

Then you draw the red one, but now there are 5 marbles in the jar (2 red ones and 3 green ones), then the probability is 2/5

The joint probability is:

P(1) = (4/6)*(2/5) = (2/3)*(2/5) = 4/15

The probability of drawing two green marbles (X = 2)

Again, first we draw a green marble with a probability of 4/6

Now we draw again a green marble, now there are 3 green marbles and 5 total marbles in the jar, so this time the probability is 3/5

Now we draw the red marble (there are 2 red marbles and 4 total marbles in the jar), with a probability of 2/4

The joint probability is:

P(2) = (4/6)*(3/5)*(2/4) = (2/6)*(3/5) = 1/5

The probability of drawing 3 green marbles (X = 3)

At this point you may already understand the pattern:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

finally, we draw a red marble with a probability 2/3

The joint probability is:

P(3) = (4/6)*(3/5)*(2/4)*(2/3) = (2/6)*(3/5)*(2/3) = (1/5)*(2/3) = (2/15)

Finally, the probability of drawing four green marbles (X = 4) is given by:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

fourth, we draw a green marble with a probability 1/3

Finally, we draw a red marble with a probability 2/2 = 1

The joint probability is:

P(4) = (4/6)*(3/5)*(2/4)*(1/3)*1 = (1/5)*(1/3) = 1/15

Then the probability distribution is:

P(0) = 1/3

P(1) = 4/15

P(2) = 1/5

P(3) = 2/15

P(4) = 1/15

The expected value will be:

EV = 0*P(0) + 1*P(1) + 2*P(2) + 3*P(3) + 4*P(4)

EV = 1*(4/15) + 2*( 1/5) + 3*( 2/15) + 4*(1/15 ) = 1.33

So we can expect to draw 1.33 green marbles in this experiment.

5 0

3 years ago

How do you solve shaded areas

marissa [1.9K]

If it is rectangular or square then just multiply the length x width and you get the area (shaded area)

4 0

3 years ago

Yall know the drill ;)

Marysya12 [62]

Answer:

-0.5 is probably right I'm not 100% sure

7 0

3 years ago

Read 2 more answers

What is the following quotient?

mamaluj [8]

Answer:

A. $2(3^{\frac{1}{3}})-\sqrt[3]{18}$

Step-by-step explanation:

We are given that

$\frac{6-3(\sqrt[3]{6}}{\sqrt[3]{9}})$

We have to find the quotient.

$\frac{6}{\sqrt[3]{9}}-3(\frac{\sqrt[3]{6}}{\sqrt[3]{9}})$

$\frac{2\times 3}{\sqrt[3]{3^2}}-3(\frac{\sqrt[3]{3\times 2}}{\sqrt[3]{3^2}})$

$2\times\frac{3}{3^{\frac{2}{3}}}-3(\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3^{\frac{2}{3}}})$

Using the property

$(ab)^n=a^n\cdot b^n$

$2\times 3^{1-\frac{2}{3}}-3(2^{\frac{1}{3}}\times 3^{\frac{1}{3}-\frac{2}{3}})$

Using the property

$\frac{a^x}{a^y}=a^{x-y}$

$2(3^{\frac{1}{3}})-3(2^{\frac{1}{3}}\times 3^{-\frac{1}{3}})$

$2(3^{\frac{1}{3}})-2^{\frac{1}{3}}\times 3^{1-\frac{1}{3}}$

$2(3^{\frac{1}{3}})-2^{\frac{1}{3}}\times 3^{\frac{2}{3}}$

$2(3^{\frac{1}{3}})-2^{\frac{1}{3}}\times \sqrt[3]{3^2}$

$2(3^{\frac{1}{3}})-2^{\frac{1}{3}}\times \sqrt[3]{9}$

$2(3^{\frac{1}{3}})-\sqrt[3]{2\times 9}$

$2(3^{\frac{1}{3}})-\sqrt[3]{18}$

Hence, the quotient of $\frac{6-3(\sqrt[3]{6}}{\sqrt[3]{9}})$ is given by

$2(3^{\frac{1}{3}})-\sqrt[3]{18}$

Option A is correct.

4 0

2 years ago

Read 2 more answers