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18 days ago

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You are a caterer specializing in children's birthday parties. You have 12 birthdays to cater next week. You must bake 2 cakes f

or each party. Each cake will have 6 candles on it. How many birthday candles do you need for the 12 parties.

1 answer:

tester [12K]18 days ago

8 0

Answer:

We require a total of 144 birthday candles for the 12 parties.

Step-by-step explanation:

As a caterer focused on children’s birthday celebrations, you have 12 events to cater to in the upcoming week, each requiring 2 cakes. Each cake will be adorned with 6 candles.

Thus, for all 12 birthday gatherings, you need a total of =

= 24 cakes.

Further, since every cake holds 6 candles, the overall number of candles for the 24 cakes amounts to =

= 144 candles.

$12 \times 2$

Therefore, you will need 144 birthday candles to accommodate the 12 celebrations.

$24 \times 6$

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A triangle has sides that measure 4 units, 6 units, and 7.21 units. What is the area of a circle with a circumference that equal

PIT_PIT [12026]

Step-by-step explanation:

Given are

Sides of the triangle measure 4 units, 6 units, and 7.21 units.

We need to compute the area of a circle whose circumference matches the triangle's perimeter.

The triangle's perimeter corresponds to the circle's circumference.

4 + 6 + 7.21 = 2πr, where r is the radius of the circle.

r = 2.73 units

Circle's area is:

$A=\pi r^2\\\\A=3.14\times (2.73)^2\\\\A=23.402\ \text{units}^2$

or

A = 24 square units

Thus, the circle’s area is 24 square units.

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1 month ago

A bag of marbles can be divided evenly among 2,3 or 4 friends. A) How many marbles might be in the bag? B) What is the least num

Leona [12263]

Response:

a)12 b)12 c)36

Detailed steps:

a) Identify the least common multiple. The quantity of marbles corresponds to that value multiplied by n (with n representing an integer)

b) The least common multiple is calculated as 4*3=...

c)12*3 =?

Afterward, simply divide by 2, 3, and 4 to find the solution

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20 days ago

The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan^−1.(t/2), where t is measured in d

AnnZ [12051]

Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

The mass of the bacterial colony (in grams) is described by

$P(t)=2+5tan^{-1}(\frac{t}{2})$

Where t=Time(in days)

Next, we differentiate with respect to t

$P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})$

Using the formula $\frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}$

$P'(t)=\frac{5}{2}(\frac{4}{4+t^2})$

$P'(t)=\frac{10}{4+t^2}$

We know that P(t)=6

Now, substitute this value

$6=2+5tan^{-1}(\frac{t}{2})$

$5tan^{-1}(\frac{t}{2})=6-2=4$

$tan^}{-1}(\frac{t}{2})=\frac{4}{5}$

$\frac{t}{2}=tan(\frac{4}{5})$

$t=2tan(\frac{4}{5})$

Insert the given value of t

$P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}$

We understand that $1+tan^2\theta=sec^2\theta$

Applying the formula

$P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})$

By employing $cos^2x=\frac{1}{sec^2x}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21$ g/day

Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

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1 month ago

Read 2 more answers

One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit

tester [12051]

Answer:

a) Robot Reliability = 0.7876

b1) Component 1: 0.8034

Component 2: 0.8270

Component 3: 0.8349

Component 4: 0.8664

b2) To maximize overall reliability, Component 4 should be backed up.

c) To achieve the highest reliability of 0.8681, backup for Component 4 with a reliability of 0.92 should be implemented.

Step-by-step explanation:

Component Reliabilities:

Component 1 (R1): 0.98

Component 2 (R2): 0.95

Component 3 (R3): 0.94

Component 4 (R4): 0.90

a) The reliability of the robot can be determined by calculating the reliabilities of the individual components that constitute the robot.

Robot Reliability = R1 x R2 x R3 x R4

= 0.98 x 0.95 x 0.94 x 0.90

Robot Reliability = 0.787626 ≅ 0.7876

b1) As only a single backup can be used at once, and its reliability matches that of the original, we evaluate each component's backup sequentially:

Robot Reliability with Component 1 backup is calculated by first assessing the failure probability of the component plus its backup:

Failure probability = 1 - R1

= 1 - 0.98

= 0.02

Combined failure probability for Component 1 and backup = 0.02 x 0.02 = 0.0004

Thus, reliability of combined Component 1 and backup (R1B) = 1 - 0.0004 = 0.9996

Robot Reliability = R1B x R2 x R3 x R4

= 0.9996 x 0.95 x 0.94 x 0.90

Robot Reliability = 0.8034

To determine reliability of Component 2:

Failure probability for Component 2 = 1 - 0.95 = 0.05

Combined failure probability of Component 2 and backup = 0.05 x 0.05 = 0.0025

Reliability of Component 2 with backup (R2B) = 1 - 0.0025 = 0.9975

Robot Reliability = R1 x R2B x R3 x R4

= 0.98 x 0.9975 x 0.94 x 0.90

Robot Reliability = 0.8270

Robot Reliability with backup of Component 3 calculates as follows:

Failure probability for Component 3 = 1 - 0.94 = 0.06

Combined failure probability of Component 3 and backup = 0.06 x 0.06 = 0.0036

Reliability for Component 3 with backup (R3B) = 1 - 0.0036 = 0.9964

Robot Reliability = R1 x R2 x R3B x R4

= 0.98 x 0.95 x 0.9964 x 0.90

Robot Reliability = 0.8349

Robot Reliability with Component 4 backup calculates as:

Failure probability for Component 4 = 1 - 0.90 = 0.10

Combined failure probability of Component 4 and backup = 0.10 x 0.10 = 0.01

Reliability for Component 4 and backup (R4B) = 1 - 0.01 = 0.99

Robot Reliability = R1 x R2 x R3 x R4B

= 0.98 x 0.95 x 0.94 x 0.99

Robot Reliability = 0.8664

b2) The best reliability is achieved with the backup of Component 4, yielding a value of 0.8664. Thus, Component 4 is the best candidate for backup to optimize reliability.

c) A reliability of 0.92 indicates a failure probability of = 1 - 0.92 = 0.08

We can compute the probability of failure for each component along with its backup:

Component 1 = 0.02 x 0.08 = 0.0016

Component 2 = 0.05 x 0.08 = 0.0040

Component 3 = 0.06 x 0.08 = 0.0048

Component 4 = 0.10 x 0.08 = 0.0080

Thus, the reliabilities for each component and its backup become:

Component 1 (R1BB) = 1 - 0.0016 = 0.9984

Component 2 (R2BB) = 1 - 0.0040 = 0.9960

Component 3 (R3BB) = 1 - 0.0048 = 0.9952

Component 4 (R4BB) = 1 - 0.0080 = 0.9920

Reliability of robot including backups for each of the components can be calculated as:

Reliability with Backup for Component 1 = R1BB x R2 x R3 x R4

= 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Backup for Component 1 = 0.8024

Reliability with Backup for Component 2 = R1 x R2BB x R3 x R4

= 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Backup for Component 2 = 0.8258

Reliability with Backup for Component 3 = R1 x R2 x R3BB x R4

= 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Backup for Component 3 = 0.8339

Reliability with Backup for Component 4 = R1 x R2 x R3 x R4BB

= 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Backup for Component 4 = 0.8681

To maximize overall reliability, Component 4 should be backed up at a reliability of 0.92, achieving an overall reliability of 0.8681.

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A boy climbs to the top of a tree and sees his friend 100 feet from the base of the tree. If the tree is 25 feet tall, what is t

Inessa [12265]

Visualizing the scenario is advantageous; creating a diagram would show a right triangle where the base measures 100 ft and the height is 25 ft. We can easily determine the angle using trigonometric functions as follows:

tan(theta) = 25 / 100
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1 month ago