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Tpy6a [65]
3 years ago
8

Log(5)=3-log(x) ????

Mathematics
1 answer:
lubasha [3.4K]3 years ago
6 0

Answer:

x = 200

Step-by-step explanation:

{10}^{ log(5) } =  {10}^{3 -  log(x) }

5 =  {10}^{3}  \times  {10}^{log( \frac{1}{x} )}

5 =  \frac{1000}{x}

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It is estimated that 30% of households own a riding lawn mower. A sample of 17 households is studied. What is the standard devia
alina1380 [7]

Answer:

1.89 is the standard deviation of households who own a riding mower

Step-by-step explanation:

Let the random variable X denote the number of households owning a riding lawn mower. From the given information, X will be a Binomial random variable with parameters; n = 17 and  p = 30% = 0.3

The standard deviation of X;

\sqrt{n*p*(1-p)}

\sqrt{17*0.3*(1-0.3)} =1.89

7 0
3 years ago
What is an example of a solution of zero ?
mart [117]
4-4=0 is an example of that
5 0
3 years ago
A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa
gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4  | red dice) = \dfrac{1}{6}

P (4 | green dice) = \dfrac{3}{6} =\dfrac{1}{2}

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = \dfrac{1}{2}

The probability of two 1's and two 4's in the first dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^4

= \dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4

= 6 \times ( \dfrac{1}{6})^4

= (\dfrac{1}{6})^3

= \dfrac{1}{216}

The probability of two 1's and two 4's in the second  dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^2  \times  \begin {pmatrix} \dfrac{3}{6}  \end {pmatrix}  ^2

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= 6 \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= ( \dfrac{1}{6}) \times  ( \dfrac{3}{6})^2

= \dfrac{9}{216}

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = \dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}

The probability of two 1's and two 4's in both die = \dfrac{1}{432}  + \dfrac{1}{48}

The probability of two 1's and two 4's in both die = \dfrac{5}{216}

By applying  Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's )  = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's )  = \dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}

P(second die (green) | two 1's and two 4's )  = \dfrac{0.5 \times 0.04166666667}{0.02314814815}

P(second die (green) | two 1's and two 4's )  = 0.9

Thus; the probability the die chosen was green is 0.9

8 0
3 years ago
Roger is paid $50 to sell videos of a play to an audience after the play is preformed.
konstantin123 [22]

Answer:

Roger can earn $510 at most.

Step-by-step explanation:

We are given the function



Which gives the earnings of Roger when he sells v videos. Since the play’s audience consists of 230 people and each one buys no more than one video, v can take values from 0 to 230, i.e.  



That is the practical domain of E(v)

If Roger is in bad luck and nobody is willing to purchase a video, v=0

If Roger is in a perfectly lucky night and every person from the audience wants to purchase a video, then v=230. It's the practical upper limit since each person can only purchase 1 video

In the above-mentioned case, where v=230, then



Roger can earn $510 at most.

3 0
3 years ago
is it possible for a percent increase or decrease to be over 100% explain why or why not this is possible
Ivahew [28]
No because there has to be a remaining opposite persent
6 0
3 years ago
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