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3 years ago

Which expression is equivalent to (2 cubed) Superscript negative 5?.

Mathematics

1 answer:

Trava [24]3 years ago

3 0

Answer:

i think its c but not sure

Step-by-step explanation:

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Mr. Clark has a tank that holds 55,000 milliliters of gas. He poured some gas into three containers equally. There were 19,864 m

Temka [501]

Answer:

35,136 ml

Step-by-step explanation:

The tank holds 55,000ml. Mr. Clark poured some gas into 3 container, and at the end there the tank had 19,864. To find how much gas was poured out of the tank it is the same as asking how much gas he poured in the 3 containers, all together. We have to find the difference, subtraction:

55,000 - 19,864 = 35,136ml were poured in the 3 containers or amount of gas poured out of the tank.

3 0

3 years ago

Let X be a Bernoulli rv with pmf as in Example 3.18. a. Compute E(X2 ). b. Show that V(X) 5 p(1 2 p). c. Compute E(X79).

spayn [35]

The Bernoulli distribution is a distribution whose random variable can only take 0 or 1

The value of E(x2) is p
The value of V(x) is p(1 - p)
The value of E(x79) is p

<h3>How to compute E(x2)</h3>

The distribution is given as:

p(0) = 1 - p

p(1) = p

The expected value of x2, E(x2) is calculated as:

$E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * (1- p) + 1^2 * p$

Evaluate the exponents

$E(x^2) = 0 * (1- p) + 1 * p$

Multiply

$E(x^2) = 0 +p$

Add

$E(x^2) = p$

Hence, the value of E(x2) is p

<h3>How to compute V(x)</h3>

This is calculated as:

$V(x) = E(x^2) - (E(x))^2$

Start by calculating E(x) using:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * (1- p) + 1 * p$

$E(x) = p$

Recall that:

$V(x) = E(x^2) - (E(x))^2$

So, we have:

$V(x) = p - p^2$

Factor out p

$V(x) = p(1 - p)$

Hence, the value of V(x) is p(1 - p)

<h3>How to compute E(x79)</h3>

The expected value of x79, E(x79) is calculated as:

$E(x^{79}) = \sum x^{79} * P(x)$

So, we have:

$E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p$

Evaluate the exponents

$E(x^{79}) = 0 * (1- p) + 1 * p$

Multiply

$E(x^{79}) = 0 + p$

Add

$E(x^{79}) = p$

Hence, the value of E(x79) is p

Read more about probability distribution at:

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5 0

3 years ago

Please answer the question correctly while showing detailed working.

kati45 [8]

Step-by-step explanation:

hey buddy it took a long time but it's ready for you

6 0

3 years ago

The number of calories Kira burns while riding a bike varies directly with the number of hours the bike is ridden. How many calo

Veseljchak [2.6K]

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3 years ago

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jok3333 [9.3K]

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3 years ago