The sale price of every item in a store is 85% of its usual price.

A movie Theater charges $6.50 for morning tickets and $8.75 for night tickets the theater sold 378 tickets for a total amount of

Xelga [282]

Answer:

The number of sold morning tickets are 168.

Step-by-step explanation:

Given,

Total number of tickets = 378

Total earned amount = $2929.50

Charge of each morning ticket = $6.50

Charge of each night ticket = $8.75

Solution,

Let the number of morning tickets be 'x'.

And also let the number of night tickets be 'y'.

So, the total number of tickets is the sum of number of morning tickets and number of evening tickets.

So, framing in equation form, we get;

$x+y=378\ \ \ \ equation\ 1$

Again, Total earned amount is the sum of number of morning tickets multiplied with charge of each ticket and number of evening tickets multiplied with charge of each ticket.

So, framing in equation form, we get;

$6.50x+8.75y=2929.50\ \ \ equation\ 2$

On multiplying equation 2 by 100 on both side, we get;

$100(6.50x+8.75y)=2929.50\times100\\\\650x+875y=292950\ \ \ \ equation\ 3$

Now multiplying equation 1 by '650', we get;

$650(x+y)=378\times650\\\\650x+650y=245700\ \ \ \ equation\ 4$

Now subtracting equation 4 from equation 3, we get;

$(650x+875y)-(650x+650y)=292950-245700\\\\650x+875y-650x-650y=47250\\\\225y=47250\\\\y=\frac{47250}{225}=210$

Now substituting the value of'y' in equation 1, we get;

$x+y=378\\\\x+210=378\\\\x=378-210=168$

Hence The number of sold morning tickets are 168.

Whereas The number of sold night tickets are 210.

5 0

3 years ago

An engineer wants to know if the average amount of energy used in his factory per day has changed since last year. The factory u

m_a_m_a [10]

Answer:

The calculated value Z = 1.8823 <1.96 at 0.05 level of significance

The average amount of energy used in his factory per day has changed since last year.

Step-by-step explanation:

Step(i):-

Given that mean of the Population = 2000 (MWH)

Given that size of the sample 'n' = 400

Given that mean of sample x⁻ = 2040 MWH

Given that the standard deviation = 425 MWH per day

Step(ii):-

Null hypothesis:H₀: μ = 2000 (MWH)

Alternative Hypothesis:H₁: μ≠ 2000( MWH)

Test statistic

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{2040-2000}{\frac{425}{\sqrt{400} } }$

Z = 1.8823

The calculated value Z = 1.8823 <1.96 at 0.05 level of significance

Final answer:-

The calculated value Z = 1.8823 <1.96 at 0.05 level of significance

The average amount of energy used in his factory per day has changed since last year.

3 0

2 years ago

2x 3/2/3 solve plsssssssss 10 points this is for my sis

JulsSmile [24]

Answer: Find the probability that sum of two numbers is 10. (iii) Find the probability that both the numbers are even. (id) Find the probability that an odd

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

The value V of a certain automobile that is t years old can be modeled by V(t) = 14,651(0.81). According to the model, when will

lakkis [162]

Answer:

a) The car will be worth $8000 after 2.9 years.

b) The car will be worth $6000 after 4.2 years.

c) The car will be worth $1000 after 12.7 years.

Step-by-step explanation:

The value of the car after t years is given by:

$V(t) = 14651(0.81)^{t}$

According to the model, when will the car be worth V(t)?

We have to find t for the given value of V(t). So

$V(t) = 14651(0.81)^{t}$

$(0.81)^t = \frac{V(t)}{14651}$

$\log{(0.81)^{t}} = \log{(\frac{V(t)}{14651})}$

$t\log{(0.81)} = \log{(\frac{V(t)}{14651})}$

$t = \frac{\log{(\frac{V(t)}{14651})}}{\log{0.81}}$

(a) $8000

V(t) = 8000

$t = \frac{\log{(\frac{8000}{14651})}}{\log{0.81}} = 2.9$

The car will be worth $8000 after 2.9 years.

(b) $6000

V(t) = 6000

$t = \frac{\log{(\frac{6000}{14651})}}{\log{0.81}} = 4.2$

The car will be worth $6000 after 4.2 years.

(c) $1000

V(t) = 1000

$t = \frac{\log{(\frac{1000}{14651})}}{\log{0.81}} = 12.7$

The car will be worth $1000 after 12.7 years.

5 0

2 years ago

Science textbooks are ordered every 10 years math textbooks are ordered every 5 years if new science and math textbooks were ord

Musya8 [376]

10 years, because In 5 years after 2008, 2013 you get new math books 5 more years you get another set of math books-2018, but 2008 plus 10 years you get science books which is also 2018 which is also when you get another set of math books. So the next year you get math books and science books shipped the same year is 2018. 2018-2008=10 so 10 years pass

8 0

2 years ago