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kondor19780726 [428]
2 years ago
11

Solve: x/x+2 + 5/x-3 = 25/x2-x-6

Mathematics
1 answer:
Paraphin [41]2 years ago
6 0

Answer:

x1=3  x2=-5

BUT x1=3 is not an answer because it doesnt respects the range on the original ecuation, so x2=-5 is the solution

Step-by-step explanation:

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Leslie decides to join a gym. She must pay a monthly fee plus a one time fee to open a membership. This situation can be modeled
omeli [17]

Step-by-step explanation:

Given function is,

y=55x+80

Where,

y = amount paid to the gym in dollars

x = number of months

So when x = 0, y =55(0)+80 = 80

i.e when number of months is 0, you have to pay a fee of $80.

This is the one time fee to open a membership.

After that for each month you have to pay $55 for each months.

So 55 is multiplied with x.

3 0
2 years ago
Can someone help me please I need the mode, median, range and the mean.?
Neko [114]

Answer:

First the mode.  Since 5 popped up the most, 5 is the mode.

Next is the median.  I crossed 1 dot from each side until it shows the last dot, and 5 was the last one.

After that the range.  9-2=7

Finally the worst, the mean... 2+2+3+3+3+4+5+5+5+5+5+6+6+6+8+9+9+9+9

=104/19=5.47

SO, Mode=5 Median=5 Range=7, and the mean is 5.47 (rounded nearest hundred)

7 0
3 years ago
Hatah and his 3 friends are each running an equal part of a la mile race.
Lana71 [14]

Answer:3.5 miles

Step-by-step explanation:

8 0
3 years ago
3m² – 17m + 10 factoring polynomials
JulsSmile [24]
Answer:

(3m-2) x (m-5)
5 0
2 years ago
1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The avera
ioda

Answer:

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

                \frac{Xbar - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = $1080

                s  = sample standard deviation = $260

                 n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, \mu is given by;

P(-2.032 < t_3_4 < 2.032) = 0.95

P(-2.032 < \frac{Xbar - \mu}{\frac{s}{\sqrt{n} } } < 2.032) = 0.95

P(-2.032 * {\frac{s}{\sqrt{n} } < {Xbar - \mu} < 2.032 * {\frac{s}{\sqrt{n} } ) = 0.95

P(X bar - 2.032 * {\frac{s}{\sqrt{n} } < \mu < X bar + 2.032 * {\frac{s}{\sqrt{n} } ) = 0.95

95% confidence interval for \mu = [ X bar - 2.032 * {\frac{s}{\sqrt{n} } , X bar + 2.032 * {\frac{s}{\sqrt{n} } ]

                                            = [ 1080 - 2.032 * {\frac{260}{\sqrt{35} } , 1080 + 2.032 * {\frac{260}{\sqrt{35} } ]

                                             = [ 990.70 , 1169.30 ]

<em>No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.</em>

Therefore, the store manager believe is not correct.

8 0
3 years ago
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