Eight friends share 64 tickets to the school carnival. Each friend has the same

-2(2x+3)=-(x+1)-2 whats x

AysviL [449]

Answer:

x=-1

Step-by-step explanation:

4 0

3 years ago

The Print Shop provides copying services to the general public. The Print Shop has equipment which is capable of making 1,800 co

steposvetlana [31]

a) The financial analysis determining the acceptance of the special order is as follows:

Data and Financial Analysis:

Capacity of photocopier per hour = 1,800 copies

Practical capacity of equipment = 7 hours

Total copies in a day = 12,600 (1,800 x 7)

Steady demand per day = 10,000

Variable cost per copy = $0.22

Service price per copy = $0.25

Profit per copy = $0.03 ($0.25 - $0.22)

Normal profit based on steady demand of 10,000 = $300 ($0.03 x 10,000)

Profit based on maximum capacity of 12,600 = $378 ($0.03 x 12,600)

Price per copy for the special customer = $0.24

Profit per copy based on special price of $0.24 = $0.02 ($0.24 - $0.22)

Profit from special customer with 5,000 pages = $100 (5,000 x $0.02)

Profit from other probable customers = $228 (7,600 x $0.03)

Lost profit for accepting special order based on maximum capacity = $50 ($378 - $228 - $100)

Lost profit if order is not accepted and demand is only 10,000 copies = $78 ($378 - $300)

Therefore, the special order at $0.24 should be accepted.

b) The largest order from customer that the shop could accept = 5,000 + $50/$0.02

= 5,000 + 2,500

= 7,500 copies

Thus, if the customer orders 7,500 copies to be made, the shop can still realize a maximum profit of $378 by accepting orders for additional 5,100 (12,600 - $7,500) copies from others.

Learn more: brainly.com/question/22599880

7 0

2 years ago

There is a triangular parking lot at the local mall. The second angle of the triangular parking lot is six more than twice as la

Travka [436]

Answer:

The measure of the three angles are;

The first angle is 62°

The second angle is 28°

The third angle is 90°

Step-by-step explanation:

The given information are;

The shape of the parking lot = Triangular

Let the angles of the triangle be given as follows

First angle = A

Second angle = B

Third angle = C

The given triangle interior angle dimensions are;

A = 6 + 2× B

C = A + B

However, we have;

A + B + C = 180° (Angle sum property for a triangle)

Therefore;

A + B + C = 180° gives;

C + C = 180° (Transitive property)

2·C = 180°

C = 180°/2 = 90°

C = 90°

However, C = A + B therefore;

90° = A + B and, A = 6 + 2 × B, we get;

A + B = 90° (Symmetric property)

6 + 2× B + B = 90° (Substitution property)

6 + 3·B = 90°

3·B = 90° - 6° = 84°

B = 84°/3 = 28°

B = 28°

From, A = 6 + 2 × B, we have;

A = 6 + 2 × 28° = 62°

A = 62°

First angle = A = 62°

Second angle = B = 28°

Third angle = C = 90°

The measure of the three angles are;

First angle is 62°

Second angle is 28°

Third angle is 90°.

8 0

3 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

2 years ago

Ricardo ran 160 minutes last week at the park. If he runs 15% more this week, how many

Ilia_Sergeevich [38]

Answer:

4356

Step-by-step explanation:

Don't be a baka

8 0

3 years ago