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3 years ago

5) where 2 coins are toased find the probability of getting:

Mathematics

1 answer:

AleksAgata [21]3 years ago

8 0

Step-by-step explanation:

Let S be the sample space

S={HH, HT, TH, TT}

n(S)= 4

A: both the coins with some place

B: both the coins with different place

C:at least one tail

D:at the most one head

1) A={TT,HH}

n(A)=2

P(G)= n(A)/n(S)

= 2/4

= 1/2

2) B ={HT, TH}

n(B) = 2

P(B) = n(B)/n(S)

= 2/4

= 1/2

3) C= {HT, TH, TT}

n(C)=3

P(C) = n(C)/n(S)

= 3/4

4). D={HT, TH}

n(D) = 2

P(D) = n(D)/n(S)

= 2/4

=1/2

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180 square yards

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3 years ago

Solve for w and simplify this fraction 3/2w=-20

AleksAgata [21]

Answer:

If I'm correct, the answer should be negative 40/3.

Step-by-step explanation:

Because you first combine multiplied terms into a single fraction, then you multiply all terms by the same value to eliminate fraction denominators. Now you simplify it, by canceling multiplied terms that are in the denominator and then you multiply the numbers. Now you will divide both sides of the equation by the same term. Finally, you will simplify the equation.

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3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

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3 years ago

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gladu [14]

I am not sure how to show you how to do a dot plot on this website, but to see how to do a dot plot

b) Put everything in order and find the middle
10, 10, 10, 20, 30, 30, 30, 30, 40, 50, 50, 50, 60, 60, 70, 70, 2040
^^
So 40 is in the middle, making it the median

c) Add all of the values together and divide by the total number of values.
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3 years ago

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saul85 [17]

Answer:

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Step-by-step explanation:

84/105=0.8

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3 years ago

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