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kicyunya [14]
3 years ago
6

Rico spent $87.00 to go to the movies on Friday night. He spent $25.75 on popcorn and drinks and paid $12.25 for each movie tick

et he bought. Which equation and solution show how many movie tickets Rico purchased on Friday night?
Mathematics
2 answers:
erastovalidia [21]3 years ago
7 0
12.25 
x=25 hope this helps
frutty [35]3 years ago
5 0
Well, you're given the values you need to create an equation. So let's take it step by step. He spent $25.75 on popcorn. Then he paid 12.25 for <em>each</em> movie ticket. We don't know how many movie tickets he bought, so we'll put x as the number of tickets for now. In total he spent, 87 dollars. 

So the equation should look like 25.75 + 12.25x = 87.00

Once you have the equation, you now have to solve for x. So let's get x by itself! 

Subtract both sides by 25.75. So you'll have 12.25x = 61.25. 

Divide both sides by 12.25, and you should get x=5, which means Rico bought 5 movie tickets.
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Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true
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Answer:

(a) 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

Step-by-step explanation:

We are given that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75.

(a) Also, the average porosity for 20 specimens from the seam was 4.85.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                      P.Q. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average porosity = 4.85

            \sigma = population standard deviation = 0.75

            n = sample of specimens = 20

            \mu = true average porosity

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

<u>So, 95% confidence interval for the true mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                     of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                            = [ 4.85-1.96 \times {\frac{0.75}{\sqrt{20} } } , 4.85+1.96 \times {\frac{0.75}{\sqrt{20} } } ]

                                            = [4.52 , 5.18]

Therefore, 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) Now, there is another seam based on 16 specimens with a sample average porosity of 4.56.

The pivotal quantity for 98% confidence interval for the population mean is given by;

                      P.Q. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average porosity = 4.56

            \sigma = population standard deviation = 0.75

            n = sample of specimens = 16

            \mu = true average porosity

<em>Here for constructing 98% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

<u>So, 98% confidence interval for the true mean, </u>\mu<u> is ;</u>

P(-2.3263 < N(0,1) < 2.3263) = 0.98  {As the critical value of z at 1% level

                                                   of significance are -2.3263 & 2.3263}  

P(-2.3263 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} <  2.3263 ) = 0.98

P( \bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.98

<u>98% confidence interval for</u> \mu = [ \bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } } ]

                                            = [ 4.56-2.3263 \times {\frac{0.75}{\sqrt{16} } } , 4.56+2.3263 \times {\frac{0.75}{\sqrt{16} } } ]

                                            = [4.12 , 4.99]

Therefore, 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

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Carter invested $16,000 in an account paying an interest rate of 5.6% compounded monthly. Assuming no deposits or withdrawals ar
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Answer:

11 years.

Step-by-step explanation:

Given that Carter invested $ 16,000 in an account paying an interest rate of 5.6% compounded monthly, to determine, assuming no deposits or withdrawals are made, how long would it take, to the nearest year, for the value of the account to reach $ 29,600, the following calculation must be performed:

16,000 x (1 + 0.056 / 12) ^ Yx12 = 29,600

Y = 11

16,000 x 1.4666 ^ 132 = X

29,581.70 = X

Thus, rounded to the nearest year, it would take 11 years for the account to reach $ 29,600.

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Using it's concept, the probability of spinning the spinner and the arrow landing on the number 1 and then on number 2 the second time is:

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<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem, we have that:

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