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timama [110]
3 years ago
7

A computer company is currently earning $400 in profit a week and predicts its profits will increase by 5% each week. A second c

ompany is currently earning $900 a week and predicts its profits will increase by 2% each week. Approximately how long will it take for the companies to make the same amount of profit?
A.
4 weeks

B.
6 weeks

C.
18 weeks

D.
28 weeks
Mathematics
1 answer:
Archy [21]3 years ago
5 0

Answer:

b 6weeks

Step-by-step explanation:

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Mnenie [13.5K]

Answer: 30 grams

Step-by-step explanation:

The interquartile range is the range within the boxed areaa. You subtract the minimum value from the maximum value.

150 - 120 = 30

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Ok guys now that you have had a good nights sleep, HElP!ME!FIGURE!THIS!OUT!10 points!
eimsori [14]

Answer:

3:5

4:4

15:7

Step-by-step explanation:

they only gave you 1 equation that is equal to 8 so all the others must be equivalent to 8

7 0
3 years ago
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For the love of God help me !! I'm desperate for it tomorrow
Eduardwww [97]
Try to relax.  Your desperation has surely progressed to the point where
you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before.  But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.

Consider this:  (2)^a negative power = (1/2)^the same power but positive.

So: 
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.

What I just said in that paragraph was:  log₂ of(N) = <em>- </em>log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.

Now let's look at the problem:

log₂(x-1) + log(base 1/2) (x-2) = log₂(x)

Subtract  log₂(x)  from each side: 

log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0

Subtract  log(base 1/2) (x-2)  from each side:

log₂(x-1) - log₂(x)  =  - log(base 1/2) (x-2)  Notice the negative on the right.

The left side is the same as  log₂[ (x-1)/x  ]

==> The right side is the same as  +log₂(x-2)

Now you have:  log₂[ (x-1)/x  ]  =  +log₂(x-2)

And that ugly [ log to the base of 1/2 ] is gone.

Take the antilog of each side:

(x-1)/x = x-2

Multiply each side by 'x' :  x - 1 = x² - 2x

Subtract (x-1) from each side:

x² - 2x - (x-1) = 0

x² - 3x + 1 = 0

Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .

I think you have to say that <em>x=2.618</em> is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.


There,now.  Doesn't that feel better. 
 






4 0
2 years ago
Can you pls help me it’s timed
Lerok [7]

Answer:

The median is 22

Step-by-step explanation:

6 0
3 years ago
A randomized controlled study was designed to test whether regular drinking of cranberry juice can prevent the recurrence of uri
Kamila [148]

Answer:

1. The chi-squared statistic = 10.36

The degrees of freedom = 17

The p-value for the test = 0.89

2. The range of the p-value from the Chi squared table = 0.75 < p-value < 0.90

Step-by-step explanation:

1. The Chi squared test is given as follows;

\chi ^{2} = \sum \dfrac{\left (Observed - Expected  \right )^{2}}{Expected  }

Therefore,

                              UTI   No UTI    %     Total

Cranberry juice       8           42      84     50

Lactobacillus          19          30       61     49

Control                    18          30      60    50

The chi-squared statistic is given as follows;

\chi ^{2} = \dfrac{\left (8- 18\right )^{2}}{18} +  \dfrac{\left (42 - 30\right )^{2}}{30} = 10.36

The chi-squared statistic = 10.36

The degrees of freedom, df = 18 - 1 = 17 since the all of the expected count have a minimum value of 18

With the aid of the calculator we find the p value as p as follows;

p = 0.9 - \dfrac{10.36 - 10.085}{12.972 - 10.085} \times (0.9 - 0.75)

The p-value for the test = 0.89  

2. The range of the p-value from the Chi squared table is given as follows;

0.75 < p-value < 0.90.

5 0
3 years ago
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