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1 year ago

How do I write log7t as a base 2 logarithm?

Mathematics

2 answers:

olga_2 [115]1 year ago

4 0

Answer:

log2t/log2^7

Step-by-step explanation:

(what the other dude said)

irina [24]1 year ago

3 0

Given:

The value is $\log_7t$ .

To find:

The value as a base 2 logarithm.

Solution:

We know that,

$\log_xy=\dfrac{\log_ay}{\log_ax}$

where, a is any positive value.

We have,

$\log_7t$

Using the above property of logarithm, we get

$\log_7t=\dfrac{\log_at}{\log_a7}$

For a=2,

$\log_7t=\dfrac{\log_2t}{\log_27}$

Therefore, the given value as a base 2 logarithm can be written as $\dfrac{\log_2t}{\log_27}$ .

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It would be 360 miles

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2 years ago

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Three-hundredths of what number is 15.3?

pychu [463]

Answer:

510

Step-by-step explanation:

0.03 × (what number) = 15.3

(what number) = 15.3/0.03 = 510 . . . . . . divide the equation by 0.03

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2 years ago

A baby otter was born 3/4 of a month early. At birth it's weight was 7/8 kilograms which is 9/10 kilogram less than the average

Assoli18 [71]

Answer:

The average weight of new born otter was, $1\frac{31}{40}$

Step-by-step explanation:

Let average weight of new born otter be x.

As per the given statement: At birth it's weight was 7/8 kilograms which is 9/10 kilogram less than the average weight of a new born otter in the aquarium

" $\frac{9}{10}$ kg less than average weight of a new born otter" means $x-\frac{9}{10}$

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$\frac{7}{8} = x -\frac{9}{10}$

Add $\frac{9}{10}$ both sides, we have;

$\frac{7}{8} + \frac{9}{10} = x$

Take LCM of 8 and 10 is, 40

⇒ $\frac{35+36}{40}=x$

Simplify:

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Therefore, the average weight of new born otter was, $1\frac{31}{40}$

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2 years ago

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For each system of equations, drag the true statement about its solution set to the box under the system?

natta225 [31]

Answer:

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

Step-by-step explanation:

* Lets explain how to solve the problem

- The system of equation has zero number of solution if the coefficients

of x and y are the same and the numerical terms are different

- The system of equation has infinity many solutions if the

coefficients of x and y are the same and the numerical terms

are the same

- The system of equation has one solution if at least one of the

coefficient of x and y are different

* Lets solve the problem

∵ y = 4x + 2 ⇒ (1)

∵ y = 2(2x - 1) ⇒ (2)

- Lets simplify equation (2) by multiplying the bracket by 2

∴ y = 4x - 2

- The two equations have same coefficient of y and x and different

numerical terms

∴ They have zero equation

y = 4x + 2

y = 2(2x - 1)

Zero solutions.

4x + 2 can never be equal to 4x - 2

∵ y = 3x - 4 ⇒ (1)

∵ y = 2x + 2 ⇒ (2)

- The coefficients of x and y are different, then there is one solution

- Equate equations (1) and (2)

∴ 3x - 4 = 2x + 2

- Subtract 2x from both sides

∴ x - 4 = 2

- Add 4 to both sides

∴ x = 6

- Substitute the value of x in equation (1) or (2) to find y

∴ y = 2(6) + 2

∴ y = 12 + 2 = 14

∴ y = 14

∴ The solution is (6 , 14)

y = 3x - 4

y = 2x + 2

One solution

3x - 4 = 2x + 2 has one solution

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2 years ago

A salesman normally makes a sale (closes) on % of his presentations. Assuming the presentations are independent, find the pro

Tamiku [17]

Answer:

$(a)\ 0.0864$

$(b)\ 0.096$

$(c)\ 0.24$

$(d)\ 0.936$

Step-by-step explanation:

Given

$p \to$ close

$q \to$ fail to close

$p = 60\%$

$p = 0.60$

First, calculate the value of q

Using complement rule

$q = 1 - p$

$q = 1 - 0.60$

$q = 0.40$

So, we have:

$p = 0.60$ and $q = 0.40$

Solving (a): Fails to close on the 4th attempt

This means that he closes the first three attempts. The event is represented as: p p p q

So, we have:

$Pr = p*p*p*q$

$Pr = p^3*q$

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$Pr = 0.0864$

Solving (b): He closes for the first time on the 3rd attempt

This means that he fails to close the first two attempts. The event is represented as: q q p

So, we have:

$Pr = q * q * p$

$Pr = q^2 * p$

$Pr = 0.40^2 * 0.60$

$Pr = 0.096$

Solving (c): First he closes is his 2nd attempt

This means that he fails to close the first. The event is represented as: q p

So, we have:

$Pr = q * p$

$Pr = 0.40 * 0.60$

$Pr = 0.24$

Solving (d): The first he close is one of his 3 attempts

To do this, we make use of complement rule

The event that he does not close any of his first three attempts is: q q q

The probability is:

$Pr = q*q*q$

$Pr = q^3$

The opposite is that the first he closes is one of the first three

So, we have:

$Pr' = 1- Pr$ --- complement rule

$Pr' = 1- q^3$

$Pr' = 1- 0.40^3$

$Pr' = 1- 0.064$

$Pr' = 0.936$

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2 years ago