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telo118 [61]
3 years ago
10

Store A sells a watch for $50 and offers a 5% discount Store B sells the same watch for $60 and offers a 20% discount. From whic

h store should you buy your
watch and why? Help please!!
Mathematics
1 answer:
erastova [34]3 years ago
6 0

Answer:

<em>You should buy the watch from Store A because it's cheaper</em>

Step-by-step explanation:

Store A sells a watch for $50 with a 5% discount.

Let's calculate the sale price. The 5% discount is

5/100*$50 = $2.50

The sale price is $50 - $2.50 = $47.50

Now for Store B. The price of the watch was $60 with a 20% discount.

Calculate 20% of $60:

20/100*$60 = $12

The sale price is $60 - $12 = $48

Thus, you should buy the watch from Store A because it's cheaper

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Textbooks are exempted from PST in Ontario. If the total cost of a book
Andrei [34K]

Answer:

4.24

Step-by-step explanation:

Total price = price of the book + 5% of the price of the book

T = total price   P = price of the book G = GST

T = P + G

and

G = 5% of P = 0.05P

therefore

T = P + 5%P = 1P + 0,05P = 1.05P

P = T/1.05 = 89/1.05 ≅ 84.76 (it is 84.7619047619)

G = 89 - 84.76 = 4.24

8 0
3 years ago
Tiva earns $48 for 6 hours of babysitting. Complete each statement if Tiva keeps earning her babysitting money at this rate. CLE
lawyer [7]
8.5
= (48 \div 6) \times 8.5
8 \times 8.5
= 68

$32

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3 0
3 years ago
Read 2 more answers
A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data
Rasek [7]

Answer: a. 1.981 < μ < 2.18

              b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

\overline{x}=\frac{\Sigma x}{n}

\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}

\overline{x}= 2.08

Now, we estimate standard deviation:

s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }

s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }

s = 0.1564

For t-score, we need to determine degree of freedom and \frac{\alpha}{2}:

df = 12 - 1

df = 11

\alpha = 1 - 0.95

α = 0.05

\frac{\alpha}{2}= 0.025

Then, t-score is

t_{11,0.025} = 2.201

The interval will be

\overline{x} ± t.\frac{s}{\sqrt{n} }

2.08 ± 2.201\frac{0.1564}{\sqrt{12} }

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

8 0
3 years ago
Help!! Which operation should be used to solve <br> X/3= 15? <br> A= ÷3<br> B= -3<br> C=+3<br> D=x3
Lera25 [3.4K]

Answer: D: X3

Step-by-step explanation:

4 0
3 years ago
What is the area of the sector with a central angle of 49° and a radius of 11 cm? Use 3.14 for pi and round your final answer to
olga_2 [115]

Answer:

\boxed{\text{51.71 cm}^{2}}

Step-by-step explanation:

If the angle θ is in radians, the formula for the area (A) of a sector of a circle is

A = ½r²θ

If θ is in degrees

A = \dfrac{1}{2}r^{2}\theta \times \dfrac{\pi \text{ rad}}{180^{\circ}}= \pi r^{2}\times\dfrac{\theta }{360}

Data:

θ = 49°

r = 11 cm

Calculation:

\begin{array}{rcl}A& = &3.14\times 11^{2}\times\dfrac{49}{360}\\\\ & = &3.14 \times 121 \times 0.1361\\ & = & \mathbf{51.71}\\\end{array}\\\text{The area of the sector is }\boxed{\textbf{51.71 cm}^{2}}

4 0
3 years ago
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