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Law Incorporation [45]
1 year ago
14

a product online for $299 and was charged an additional $10 for shipping and handling. The customer received a rebate of $99 and

a $25 credit for turning in an old cell phone. After a tax of 4% was applied, how much did the customer pay for the product?
Mathematics
2 answers:
sammy [17]1 year ago
8 0
The customer payed $192.4 for the product.
ratelena [41]1 year ago
6 0

Answer:

Amount of money paid by the customer = $192.40

Step-by-step explanation:

Online price of the product = $299

Shipping and handling charge = $10

Rebate received = $99

Credit earned for turning in an old phone = $25

So, Total amount of mobile phone = 299 + 10 - 99 - 25

                                                          = $185

Tax applied on final price = 4%

Amount of money paid by the customer = 185 + 4% of 185

= 185 + 0.4 × 185

= $192.40

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P=100a÷t solve for a
pishuonlain [190]

<u><em>a=1/100pt is the correct answer.</em></u> First you had to multiply by t from both sides of equation, and it gave us, ⇒ pt=100a. And then you had to flip an equation, and it gave us, ⇒100a=pt. You can also divide by one-hundred (100) from both sides of equation, and it gave us, \frac{100a}{100}=\frac{pt}{100}. And finally, and it gave us the answer is a=1/100pt is the final answer. Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie

8 0
2 years ago
Read 2 more answers
Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and
lubasha [3.4K]

This question is incomplete, the complete question is;

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and preview ads before the movie starts. Many complain that the time devoted to previews is too long (The Wall Street Journal, October 12, 2012). A preliminary sample conducted by The Wall Street Journal showed that the standard deviation of the amount of time devoted to previews was 4 minutes. Use that as a planning value for the standard deviation in answering the following questions. Round your answer to next whole number,

a) If we want to estimate the population mean time for previews at movie theaters with a margin of error of 75 seconds, what sample size should be used? Assume 95% confidence.

b) If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used? Assume 95% confidence.

Answer:

a) the required Sample Size is 40

b) the required Sample Size is 62

Step-by-step explanation:

Given the data in the question;

standard deviation σ = 4 minutes

a)

margin of error E = 75 seconds = ( 75 / 60 )minutes = 1.25 minutes

And 95% confidence.

Now, Critical Value of z for 0.95 confidence interval;

∝ = 1 - 0.95 = 0.05

Z_{\alpha /2 = 1.96

so, sample size n will be;

n = [ Z_{\alpha /2 × σ/E ]²

we substitute;

n = [ 1.96 × (4/1.25) ]²

n =  [ 1.96 × 3.2 ]²

n = [ 6.272 ]²

n = 39.338

Since we referring to a number of sample, its approximately becomes 40

Therefore, the required Sample Size is 40

b)

margin of error E = 1 minutes

And 95% confidence.

Now, Critical Value of z for 0.95 confidence interval;

∝ = 1 - 0.95 = 0.05

Z_{\alpha /2 = 1.96

so, sample size n will be;

n = [ Z_{\alpha /2 × σ/E ]²

we substitute;

n = [ 1.96 × (4/1) ]²

n =  [ 1.96 × 4 ]²

n = [ 7.84 ]²

n = 61.466

Since we referring to a number of sample, its approximately becomes 62

Therefore, the required Sample Size is 62

6 0
1 year ago
A group of kids just finished trick-or-treating. The number of pieces of candy collected by each of the 5 kids
NemiM [27]

Answer:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

s =\sqrt{\frac{(31-35)^2 +(33-35)^2 +(36-35)^2 +(41-35)^2 +(34-35)^2}{5-1}} =3.808

And the answer for this case would be :

s= 3.81 after round the value

Step-by-step explanation:

We have the following data set given:

31, 33, 36, 41, 34

If we want to find the standard deviation we need to find first the sample mean with this formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And replacing we got:

\bar X = \frac{31+33+36+41+34}{5}= \frac{175}{5}= 35

Now we can find the sampel deviation with this formula:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And replacing we got:

s =\sqrt{\frac{(31-35)^2 +(33-35)^2 +(36-35)^2 +(41-35)^2 +(34-35)^2}{5-1}} =3.808

And the answer for this case would be :

s= 3.81 after round the value

6 0
2 years ago
The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with me
lisov135 [29]

Answer:

a) <u>0.4647</u>

b) <u>24.6 secs</u>

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = <u>0.4647</u>

Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>

<u></u>

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = <u>24.6 secs</u>

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>

6 0
1 year ago
(14.14) larry reads that 1 out of 6 eggs contains salmonella bacteria. so he never uses more than 5 eggs in cooking. if eggs do
melomori [17]

we know that

The probability that "at least one" is the probability of exactly one, exactly 2, exactly 3, 4 and 5 contain salmonella.

The easiest way to solve this is to recognise that "at least one" is ALL 100% of the possibilities EXCEPT that none have salmonella.

If the probability that any one egg has 1/6 chance of salmonella

then

the probability that any one egg will not have salmonella = 5/6.

Therefore

for all 5 to not have salmonella


= (5/6)^5 = 3125 / 7776

= 0.401877 = 0.40 to 2 decimal places


REMEMBER this is the probability that NONE have salmonella


Therefore

the probability that at least one does = 1 - 0.40

= 0.60


the answer is

0.60 or 60%

6 0
2 years ago
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