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Dmitry_Shevchenko [17]
3 years ago
15

Mikes movie Rentals Charge $3.00 for each rental plus $0.50 for each additional day. Reed's Rentals Charge $1.50 each rental plu

s $1.25 for each additional day. For what number of additional days will the bill at both stores be the same
Mathematics
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

2

Step-by-step explanation:

What we need here is to express both charges trough equations, and then find out for which value these equations are equal.

Let X be the amount of additional days of rent.

So, Mikes Movie Rentals cost equations is:

CostMike = 3 + 0.5X

This implies that if there is no additional day (X=0), the cost is $3, if there is one additional day the cost is $3.5, for 2 days $4, and so on ...

Now lets find Reed's Rental cost.:

CostReed = 1.5 + 1.25X

And the interpretation is analogous to the one above. If there is no additional day, Reed's charge is $1.5, for 1 days $2.75, for 2 days $4

(Notice that with the examples I made about interpreting the equations I found out that for 2 days the cost of both is the same. Now lets find it analytically)

For finding the additional days that give the same costs we need to equal both equations:

3 + 0.5 X = 1.5 + 1.25 X

Lets subtract 1.5 from both sides:

3 - 1.5 + 0.5 X = 1.5 - 1.5 + 1.25 X

1.5 + 0.5 X = 1.25 X

Now subtract 0.5 X in both sides:

1.5 + 0.5 X - 0.5 X = 1.25 X - 0.5 X

1.5 = 0.75 X

Dividing both sides by 0.75:

1.5 / 0.75 = 0.75 X / 0.75

2 = X

So, for 2 additional days bith charges are equal.  

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A fabric manufacturer believes that the proportion of orders for raw material arriving late isp= 0.6. If a random sample of 10 o
ryzh [129]

Answer:

a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

Step-by-step explanation:

Given the data in the question;

proportion p = 0.6

sample size n = 10

binomial distribution

let x rep number of orders for raw materials arriving late in the sample.

(a) probability of committing a type I error if the true proportion is  p = 0.6;

∝ = P( type I error )

= P( reject null hypothesis when p = 0.6 )

= ³∑_{x=0 b( x, n, p )

= ³∑_{x=0 b( x, 10, 0.6 )

= ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.6)^x( 1 - 0.6 )^{10-x

∝ = 0.0548

Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

the probability of committing a type II error for the alternative hypotheses p = 0.3

β = P( type II error )

= P( accept the null hypothesis when p = 0.3 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.3 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - 0.6496

= 0.3504

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

the probability of committing a type II error for the alternative hypotheses p = 0.4

β = P( type II error )

= P( accept the null hypothesis when p = 0.4 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.4 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - 0.3823

= 0.6177

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

the probability of committing a type II error for the alternative hypotheses p = 0.5

β = P( type II error )

= P( accept the null hypothesis when p = 0.5 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.5 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - 0.1719

= 0.8281

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

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Step-by-step explanation:

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