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3 years ago

By definition, empirical probability is equal to:

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

4 0

Answer:

C.) Number of successful trials / Total number of trials

Step-by-step explanation:

The empirical probability of an event 'E' is defined as :

(Number of successful occurrences of event 'E' / Total number of trials)

The empirical probability requires that an experiment of a chosen number of trials be performed. The ratio of the number of success to the total number of trials is used to estimate the empirical probability of such event.

P(Empirical) = (Number of successful trials / Total number of trials)

This is different from the use of sample space in the evaluation of theoretical probability whereby all possible occurences is used to estimate the probable outcome of an event.

P(theoretical) = (number of required outcomes / total number of possible outcomes)

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Solve. 5x - 5 = 50 A) x = 8 B) x = 9 C) x = 10 D) x = 11

Angelina_Jolie [31]

The answer would be D) x = 11.

Add 5 to both sides (5x = 50 + 5)

Simplify 50 + 5 to 55 (5x = 55)

Divide both sides by 5 (x = 55/5)

Simplify 55/5 to 11 (x = 11)

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3 years ago

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Vilka [71]

Answer:

x = 0 1 -1 -2

y = -3 -4 -2 -1

Step-by-step explanation:

equation : 3y+9x = -9

when x = 1

3y+3x = -9

or, 3y + 3×1= -9

or, 3y+3=-9

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Again

Let's suppose x as -1

3y+3x= -9

or, 3y+ 3×-1 = -9

or, 3y-3 = -9

or, 3y = -6

y = -2

Let's suppose x as -2 now,

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5 0

2 years ago

The sum of four consecutive terms of an arithmetic progression is 32 and the ratio of product of the first and the last term to

OverLord2011 [107]

The four consecutive terms are 2,6,10,14.

Explanation:

Let the four consecutive terms of an arithmetic progression be (a - 3d), (a - d),(a + d), (a + 3d)

It is given that the sum of four consecutive terms is 32.

Thus, adding all the four consecutive terms, we have,

$\begin{array}{r}{a-3 d+a-d+a+d+a+3 d=32} \\{4 a=32} \\{a=8}\end{array}$

Thus, the value of a is 8.

It is also given that the ratio of the product of first and last term to the product of the two middle terms is 7:15. Thud, we have,

$\frac{(a-3d)(a+3d)}{(a-d)(a+d)} =\frac{7}{15}$

Simplifying the terms, we have,

$\begin{aligned} \frac{a^{2}-9 d^{2}}{a^{2}-d^{2}} &=\frac{7}{15} \\ 15\left(a^{2}-9 d^{2}\right) &=7\left(a^{2}-d^{2}\right) \\ 15 a^{2}-135 d^{2} &=7 a^{2}-7 d^{2} \\ 8 a^{2} &=128 d^{2} \end{aligned}$

Now, substituting a = 8, we have,

$\begin{aligned} 8(64) &=128 d^{2} \\ 512 &=128 d^{2} \\ 4 &=d^{2} \\ 2 &=d \end{aligned}$

Thus, the value of d is 2.

Now, we shall substitute the value of a and d in the terms (a - 3d), (a - d),(a + d), (a + 3d), we get, the four consecutive terms.

Thus, the four consecutive terms are 2,6,10,14.

8 0

3 years ago