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dezoksy [38]
3 years ago
15

Pampered Pooch charges $5 for one hour of Doggy Daycare $10 for two hours and fifteen dollars for 3 hours what is the rate chang

e
Mathematics
1 answer:
DochEvi [55]3 years ago
4 0

Answer:

$5 difference in rate at each step or change in hours.

Step-by-step explanation:

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Moneysaver's Bank offers a savings account that 7interest compounded continuouslyIf Donna deposits 2200, how much will she have
OverLord2011 [107]

For an initial amount deposited P and an annual interest rate r, the total amount T in the account after t years is given by:

T=P(1+r)^t

For P = $2200, r = 7% and t = 5 years, we have:

\begin{gathered} T=2200(1+0.07)^5 \\ T=\text{ \$3,085.61} \end{gathered}

3 0
11 months ago
If you minus a positive number from a negative number is the diffirance a positive number, or a negative number? For example: 5
Reika [66]

Answer:

positive + positive is negative

negative+ positive = negative ii think

Step-by-step explanation:

3 0
2 years ago
A hotel has number of meeting rooms m available for events each meeting room. Has 325 chairs write equation to represent c the t
BabaBlast [244]

Answer:

2,275

Step-by-step explanation:

c=325m

Because for every meeting room there is 325 chairs and c=the total of chairs

and c=325(7) so c=2,275

so the answer is 2,275

3 0
2 years ago
Which steps are needed to solve this equation? 100 points for the first correct answer.
GrogVix [38]

Answer:

the first one sub 6 from both =15

Step-by-step explanation:

b+6=21

-6 -6

b=15

8 0
3 years ago
In a student body, 50% use Google Chrome, 9% use Internet Explorer, 10% Firefox, 5% Mozilla, and the rest use Safari. In a group
Likurg_2 [28]

Answer:

The probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome is 0.1350.

Step-by-step explanation:

Denote the events as follows:

<em>C</em> = a student uses Google chrome

<em>E</em> = a student uses Internet explorer

<em>F</em> = a student uses Firefox

<em>M</em> = a student uses Mozilla

<em>S</em> = a student uses Safari

Given:

P (C) = 0.50

P (E) = 0.09

P (F) = 0.10

P (M) = 0.05

P (S) = 0.26

A sample of <em>n</em> = 5 students is selected.

The probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome is:

P (E = 1 ∩ C ≥ 3) = P (E = 1 ∩ C = 3) + P (E = 1 ∩ C = 4) - P (E = 1 ∩ C = 5)

The probability distribution of a student using any of the browser is Binomial.

Compute the probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome as follows:

P (E = 1 ∩ C ≥ 3) = P (E = 1 ∩ C = 3) + P (E = 1 ∩ C = 4) - P (E = 1 ∩ C = 5)

                          = P (E = 1) [P (C = 3) + P (C = 4) - P (C = 5)]

       ={5\choose 1}(0.09)^{1}(1-0.09)^{5-1}[{5\choose 3}(0.50)^{3}(1-0.50)^{5-3}+{5\choose 4}(0.50)^{4}(1-0.50)^{5-4}\\-{5\choose 5}(0.50)^{5}(1-0.50)^{5-5}]\\=0.3086[0.3125+0.1563-0.0313]\\=0.3086\times 0.4375\\=0.1350

Thus, the probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome is 0.1350.

5 0
3 years ago
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