Ask question

Ask question

All categories

3 years ago

13

The current value of an investment is 25% more than its initial value. The increase in value is $10 million. What was the initia

l value of the investment
A.12.5 million
B.40 million
C.35 million
D.100 million

1 answer:

katrin [286]3 years ago

7 0

B is the answer ////////

You might be interested in

Nine percent of americans say they are well informed about politics in comparison to most people. You randomly sample 200 americ

Arada [10]

Answer:

Step-by-step explanation:

Using normal distribution

z = (x - mean) / standard deviation

Nine percent of americans say they are well informed about politics in comparison to most people.

That means probability of success is p =9/100 = 0.09,

then probability of failure is

q = 1-p =1 -0.09= 0.91

n = number of sample = 200

x=(8% of 200) = 0.08×200 =16

mean = np = 200×0.09= 18

Standard deviation = npq = 200×0.09×0.91= 16.38

We are looking for the probability that less than 8% of the people sampled will answer yes to the question

8% of 200 = 16

P (x greater than 16) =P(x lesser than/equal to 15)

z = 15-18 /16.38 = -3/16.38

= -0.18

Looking at the normal distribution table

P(x greater than 16) = 0.5714

5 0

3 years ago

It takes 124 minutes to lift 4 tires on to a lorry how long would it take him to fit 6 tires in a lorry

Blizzard [7]

Answer:186

Step-by-step explanation:

you divide 124 by 2 and you just add the half to 124.

6 0

2 years ago

I’m wondering how this brainly

Anni [7]

Answer: i dont think so

i mean there arent many bots ive seen at least

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

2 years ago

Find the slope of the line that passes through (50, -70) and (77, 62). Simplify your answer and write it as a proper fraction, i

-Dominant- [34]

Answer:

44/9 or 4.8888888888889

Step-by-step explanation: ¯\_(ツ)_/¯

4 0

3 years ago

Read 2 more answers