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3 years ago

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At the movie theater child admission is $5.90 and adult is $9.00. Wednesday 145 tickets were sold for a total sales of 1115.90.

How many adult tickets were sold that day?

1 answer:

Anettt [7]3 years ago

8 0

Answer:1160.3

1169.30-9.00

Step-by-step explanation:1169.30-9.00

You might be interested in

A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of

GenaCL600 [577]

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.989.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

$M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23$

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

8 0

3 years ago

Please help thank you!

Arte-miy333 [17]

Answer:

golf: $10
batting: $2.50

Step-by-step explanation:

We can write two equations in the two unknowns using the given relations. Let g and b represent the costs of a round of golf and a turn in the batting cage, respectively.

5g +4b = 60 . . . . . Sylvester's expense

3g +6b = 45 . . . . . Lin's expense

Dividing the second equation by 3 gives ...

g +2b = 15 ⇒ 2b = 15 -g

Substituting into the first equation, we have ...

5g +2(2b) = 60

5g +2(15 -g) = 60 . . . . . substitute for 2b

3g = 30 . . . . . . . . . subtract 30, collect terms

g = 10 . . . . . . . divide by 3

__

2b = 15 -10 = 5 . . . . use the value of g to find b

b = 2.5 . . . . . . . . divide by 2

Mini golf costs $10 per round; batting cages cost $2.50 per turn.

7 0

2 years ago

Find the missing side length to the nearest tenth of a millimeter.

Zarrin [17]

Answer:

A = 4.7

Step-by-step explanation:

To find the missing leg, use this formula:

$a = \sqrt{c^2 - b^2}$

----------------------------------------------------------------------------------------------------------------

Have a good day :)

5 0

3 years ago

What is the y-intercept of the function f(x)=4 - 5x? (Hint: use the commutative property)

Tom [10]

Answer:

Therefore the y-intercept of the function is 4.

Step-by-step explanation:

Intercepts:

The line which intersect on x-axis and y-axis are called intercepts.

y-intercept: The line or function which intersect at y-axis. So when the line intersect at y-axis, X coordinate is zero.

So in the given Function Put x = 0 we will get the y-intercept

$f(x)= 4-5x$

Put x =0

$f(0)= 4-5\times 0$

$f(0)= 4-0=4$

Therefore the y-intercept of the function is 4.

7 0

3 years ago

I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.

masya89 [10]

Answer:

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

$P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.$

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

$P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.$

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

4 0

2 years ago