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3 years ago

If 3 people stay in a hotel for today and by gh600.00 .what is the cost of 5 people staying for 8 days.

1 answer:

3 0

so it'd be 600 / 3 which gives you 200.00 that's what each person paid a day given that there's 5 five times 200 is one day 1000.00x 8 = ($8000.00) for 5 people for 8 days

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Seven minus 2 times a number is the same as the number minus 2

Ronch [10]

Assume the number is x
now, we will translate the written english into equation:
seven minus 2 times the number is 7-2x
the number minus 2 is x-2
we will equate these two equations (the question says that they are equal) ans solve for x as follows:
7-2x=x-2
7+2=x+2x
9=3x
x=3
the required number is 3

7 0

3 years ago

Read 2 more answers

When baking a cake, you have a choice of the following pans:

Lorico [155]

Answer:

It is 6x9

Step-by-step explanation:

Simple match 9x6 54 and 7x2 14 well I would rather have 54 units of cake that 14 Lol

8 0

3 years ago

Determine the rate of change for 3/2x -5/4y = 15

Aleksandr-060686 [28]

Answer:

-6/5

Step-by-step explanation:

3/2x - 5/4y = 15

-5/4y = -3/2x + 15

multiply each side by (-4/5)

(-4/5)(-5/4y) = (-4/5)(-3/2x + 15)

y = -6/5x - 12

rate of change: -6/5

6 0

3 years ago

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago

A laptop company claims up to 11.0 hours of wireless web usage for its newest laptop battery life. However, reviews on this lapt

emmainna [20.7K]

Answer:

Step-by-step explanation:

given that a laptop company claims up to 11.0 hours of wireless web usage for its newest laptop battery life. However, reviews on this laptop shows many complaints about low battery life. A survey on battery life reported by customers shows that it follows a normal distribution with mean 10.5 hours and standard deviation 27 minutes.

convert into same units into hours.

X is N(10.5, 0.45)

a) the probability that the battery life is at least 11.0 hours

$P(X\geq 11)\\\\=1-0.8667\\=0.1333$

(b) the probability that the battery life is less than 10.0 hours

= $P(X$

=97th percentile

= 11.844

d) The time of use that is exceeded with probability 0.9 is

is 90th percentile = 10.885

8 0

3 years ago