All categories

2 years ago

How to change base of logarithms: Best answer gets 27 points and brainliest answer

Mathematics

1 answer:

AlekseyPX2 years ago

6 0

Answer:

1/3

Step-by-step explanation:

To change from one base to another, we use the formula

Logb x = Loga x/Loga b

log1/9 (3^(1/3) /3)

log3 ((3^(1/3) /3))

-------------------------

log3 (1/9)

Log a /b = log a - log b

and 1/9 = 3^-2

log3 ((3^(1/3) ) - log3 (3)

-------------------------

log3 (3^-2)

log a^b = blog a

1/3 log3 (3 ) - log3 (3)

-------------------------

-2log3 (3)

We know log3 (3) =1

1/3 (1) - 1

-------------------------

-2 (1)

1/3 - 1

-------------------------

-2

-2/3

------

-2

Copy dot flip

-2/3 * -1/2

1/3

You might be interested in

5 points

azamat

Step-by-step explanation:

it so easy bro 5.5 ok bro.

3 0

2 years ago

PLEASSEEEE HELP W THIS QUESTION!!!!

brilliants [131]

X/28 = 5/7
x = 28 * 5 / 7
x = 20

answer
x = 20 ft

3 0

2 years ago

Read 2 more answers

In a particular game, a fair die is tossed. If the number of spots showing is either four or five, you win $1. If the number of

TiliK225 [7]

Answer:

The probability that you win at least $1 both times is 0.25 = 25%.

Step-by-step explanation:

For each game, there are only two possible outcomes. Either you win at least $1, or you do not. Games are independent. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of winning at least $1 on a single game:

The die has 6 sides.

If it lands on 4, 5 or 6(either of the three sides), you win at least $1. So

$p = \frac{1}{2} = 0.5$

You are going to play the game twice.

This means that $n = 2$

The probability that you win at least $1 both times is

This is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{2,2}.(0.5)^{2}.(0.5)^{2} = 0.25$

The probability that you win at least $1 both times is 0.25 = 25%.

4 0

2 years ago

Pls help me with this

scoundrel [369]

Answer:

6mm

Step-by-step explanation:

7 0

1 year ago

I bought my flat last year for £40,000. The value of my flat increased by 12% in September and then by a further 10%in January t

Leya [2.2K]

The answer is: <span>40,000(1.12)(1.10)

Or 4 more understanding: 12% of £40000
</span>1%= 400, therefore, 12%= 4800.
In January, the value was £44800.
10% of this is £448.
Add £448 to £44800, giving you £45, 248

7 0

2 years ago