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Nutka1998 [239]

3 years ago

9

Whats the answer

le="80 + ( - 22 \frac{4}{15})" alt="80 + ( - 22 \frac{4}{15})" align="absmiddle" class="latex-formula">

1 answer:

yaroslaw [1]3 years ago

4 0

The answer is =-214/15

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Diego wants to have $1,000,000 in 5 years. He plans to invest $250,000 to start and make yearly payments of $125,000 to the acco

julia-pushkina [17]

Answer:

Yes

Step-by-step explanation:

With the $875,000 from his input alone he is really close to $1,000,000 and doing simple intrest on the $250,000 is $105,000 a year is he earns 3.5% intrest a month.

So in 5 years he'll have about $1,662,500 using simple intrest but with compound it'd be more like 2mil

Hope this helps!

5 0

2 years ago

Which pair of complex factors results in a real-number product?

Tcecarenko [31]

Answer:

1l

Step-by-step explanation:

5n

6 0

3 years ago

A box in a supply room contains 18 compact fluorescent lightbulbs, of which 6 are rated 13-watt, 8 are rated 18-watt, and 4 are

V125BC [204]

Answer:a) Probability P(exactly 2 bulbs rated 13watts)= 0.22

b) Probability(each bulb different rating)

= 0.24

Step-by-step explanation:

There are 6 13watts bulbs

8 18watts bulbs

4 23watts bulbs

Total bulbs = 18

a) Probability that all 3 bulbs are 18watts

Number of ways of pulling 3 bulbs = 18!/(3!×15!) = (6.4×10^15)/7.846×10^12) = 815 ways

Different ways of pulling 13watts bulbs out of 6 = 6!/(2!×4!)= 720/48=15ways

Different ways of pulling non 13 watts bulbs= 12!/(1!×11!) = 479,001,600/ 39,916,800 = 12

Number of ways total= 15×12=180waya

Therefore P(exactly 2 bulbs rated 13watts)= 180/815 =0.22

b) Probability P( all 3 bulbs are 1 from each rating)

Ways of pulling 3bulbs bulb each from 3 ratings are 4× 5 × 8= 192ways

Probability = 192/815 =0.24

6 0

3 years ago

Read 2 more answers

If a light bulb manufacturing company wants to estimate, with 99% confidence, the mean life of compact fluorescent light bulb

Reptile [31]

Answer:

We need at least 217 compact fluorescent light bulbs

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How many compact fluorescent light bulbs need to be selected?

We need at least n bulbs, in which n is found when $M = 175, \sigma = 1000$

So

$M = z*\frac{\sigma}{\sqrt{n}}$

$175 = 2.575*\frac{1000}{\sqrt{n}}$

$175\sqrt{n} = 2575$

$\sqrt{n} = \frac{2575}{175}$

$\sqrt{n} = 14.71$

$\sqrt{n}^{2} = (14.71)^{2}$

$n = 216.5$

We need at least 217 compact fluorescent light bulbs

4 0

3 years ago

Read 2 more answers

Nathan jogged 2 /1 4 m i l e s o n M o n d a y . W e d n e s d a y h e j o g g e d 3 /1 3 m i l e s , a n d F r i d a y , h e j

jeka57 [31]

Answer: 8 1/4 miles

Step-by-step explanation:

I think your question isn't well written and should be:

Nathan jogged 2 1/4 miles on Monday. On Wednesday, he jogged 3 1/3 miles and on Friday, he jogged 2 2/3 miles. How far did Nathan jog altogether?

This will be:

= 2 1/4 + 3 1/3 + 2 2/3

= 2 3/12 + 3 4/12 + 2 8/12

= 7 15/12

= 7 + 1 3/12

= 8 3/12

= 8 1/4 miles

3 0

3 years ago