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4 years ago

Which logarithmic equation is equivalent to the exponential below? e^a= 35

Mathematics

1 answer:

lara [203]4 years ago

8 0

Answer:

$log_{e}$ 35 = a

Step-by-step explanation:

using the law of logarithms

• $log_{b}$ x = n ⇔ x = $b^{n}$

$e^{a}$ = 35 ⇒ $log_{e}$ 35 = a

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Glenda checks the computer monitors made at her company for defects. It has been shown that 0.5% of the monitors are defective.

Svetllana [295]

The statement that will make this process into a binomial experiment is that "C. Glenda checks the monitors until she has 300 non-defective monitors."

Here are the following choices:
<span>A. Glenda checks the monitors until she has 300 defective monitors.
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C. Glenda checks the monitors until she has 300 non-defective monitors.
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3 years ago

A triangle with side lengths of 5, 12, and

nikklg [1K]

Answer:

51.2

Step-by-step explanation:

$\frac{15}{5} = \frac{25}{x}$

15x = 120

x = 8

so now we know that 2 sides of the bigger triangle are 24 and 8

now

$\frac{8}{x} = \frac{5}{12}$

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4 years ago

9/2 - 23/9=? plss help

Galina-37 [17]

Answer:

35/ 18

(Decimal: 1.944444)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Help me wit dis plz for for 20 points

Alisiya [41]

Answer:

I dont know but combine like terms and then solve it will make It alot easier

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3 years ago

If discount rate is 12%, the present value of Rs X recieved at the end of each year for the next five years is equal to

myrzilka [38]

Answer:

Present Value = $X [\frac{1}{(1 + 0.12)^{1} } + \frac{1}{(1 + 0.12)^{2} } + \frac{1}{(1 + 0.12)^{3} } + \frac{1}{(1 + 0.12)^{4} } + \frac{1}{(1 + 0.12)^{5} } ]$

Step-by-step explanation:

To find - If discount rate is 12%, the present value of Rs X received at the end of each year for the next five years is equal to .... ?

Solution -

We know that, formula for finding the Present vale is given by

Present value = Future value / (1 + r)ⁿ

where r is the rate of interest

and n is Number of periods

Now,

Here in the question, we have

r = 12% = 12/100 = 0.12

n = 5

Also, Given that, we have received Rs X at the end of each year

So,

Present Value = $\frac{X}{(1 + 0.12)^{1} } + \frac{X}{(1 + 0.12)^{2} } + \frac{X}{(1 + 0.12)^{3} } + \frac{X}{(1 + 0.12)^{4} } + \frac{X}{(1 + 0.12)^{5} }$

= $X [\frac{1}{(1 + 0.12)^{1} } + \frac{1}{(1 + 0.12)^{2} } + \frac{1}{(1 + 0.12)^{3} } + \frac{1}{(1 + 0.12)^{4} } + \frac{1}{(1 + 0.12)^{5} } ]$

⇒Present Value = $X [\frac{1}{(1 + 0.12)^{1} } + \frac{1}{(1 + 0.12)^{2} } + \frac{1}{(1 + 0.12)^{3} } + \frac{1}{(1 + 0.12)^{4} } + \frac{1}{(1 + 0.12)^{5} } ]$

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3 years ago