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2 years ago

A question that is used as the focus of a PBL project that communicates the challenge you must solve is called __________.

1 answer:

6 0

<u>Project-Based Learning</u> is an educational approach based on student-centered instruction that concerns an active classroom technique, understanding and aptitudes by performing for an extended period to study and react to a genuine, fascinating, and difficult question or challenge.

A question used for focusing on a PBL task that expresses the challenge you must unravel is a <u>driving question</u>.

<h3>What do you mean by the driving question?</h3>

A driving question is the principle of a project-based learning exercise. A driving challenge is open-ended and functions as the priority of the task.

It is connected to the central purpose of the task and the objective assignments that the instructor wants the learners to profit from.

Therefore, the driving question is the correct blank.

Learn more about project-based learning here:

brainly.com/question/26265333

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Which of the following is true about a business plan?

dusya [7]

Answer:

Explanation:

the plan can change from buget sponsors and othere factors

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3 years ago

_____________ are a form of financial instrument through which corporations and governments borrow money from financial investor

makkiz [27]

Answer:

Bonds

Explanation:

Bonds are financial instruments that are used to obtain funding from the bond holders. It is a debt security that is issued by a government or corporation to investors.

When investors buy bonds the funds are used by governments for its operations and various projects. Interest is paid on the bonds.

Bonds can be municipal bonds or corporate bonds.

Unlike shares bonds does not grant the holder an equity or ownership stake in the company, rather it grants a creditor stake.

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2 years ago

A company purchased 300 units for $60 each on January 31. It purchased 150 units for $25 each on February 28. It sold a total of

Daniel [21]

Answer:

Weighted-average inventory costing method Ending Inventory = $ 9666.67= $ 9667

Explanation:

Date Particulars Units Unit Cost Total Cost

January 31 Purchases 300 $ 60 $ 18,000

February 28 Purchases 150 $ 25 $3750

Total 450 $ 21,750

Weighted-average inventory costing method= Total Cost/ Total Units=

$ 21,750/450= $48.33 purchase price per unit

Sales 250 units at $ 70 = $ 17500

Ending Units = Purchases-Sales = 450-250= 200

Weighted-average inventory costing method Ending Inventory = $ 9666.67

200 units at 448.33= $ 9666.67= $ 9667

3 0

3 years ago

Read 2 more answers

Baxter Inc. owns 90 percent of Wisconsin Inc. and 20 percent of Cleveland Company.

dexar [7]

Answer:

$350,380

Explanation:

Calculation to determine the amount that would appear on the consolidated income

Consolidated income statement.

Sales$1,590,000

($1,000,000+$450,000+$280,000-$100,000-$40,000)

Less :Cost of goods sold ($1,015,000)

($670,000+$280,000+$190,000-$100,000-$25,000)

Less :Expenses ($200,000)

($110,000+$60,000+$30,000)

Dividend income$0

Consolidated net income $375,000

Noncontrolling interests in subsidiaries' income $24,620

Controlling interest in consolidated net income $350,380

Therefore the amount that would appear on the consolidated income will be $350,380

8 0

3 years ago

Repair calls are handled by one repairman at a photocopy shop. Repair time, including travel time, is exponentially distributed,

wolverine [178]

Answer:

the average number of customers awaiting repairs = 0.30

the system utilization = 42

the amount of time that the repairman is not out on a call is = 4.64 hours

the probability of two or more customers in the system = 0.1764

Explanation:

Given that :

Repair time, including travel time = mean of 1.6 hours per call.

Requests for copier repairs = mean rate of 2.1 per eight-hour day

i.e mean rate R = 2.1/day

Time = 8 hours

thus; mean rate μ = 8 hours/ 1.6 hours = 5

(a)

Let the average number of customers awaiting repairs be $I_i$ :

$I_i = \dfrac{R^2}{\mu (\mu-R)}$

$I_i = \dfrac{2.1^2}{5 (5-2.1)}$

$I_i = \dfrac{4.41}{5 (2.9)}$

$I_i = \dfrac{4.41}{14.5}$

$\mathbf{I_i = 0.30}$

the average number of customers awaiting repairs = 0.30

(b) Determine system utilization.

The system utilization is determined as follows:

$\delta = \dfrac{R}{\mu}$

$\delta = \dfrac{2.1}{5}$

${\delta = 0.42}$

$\mathbf{\delta = 42}$

Percentage of Idle time = 1 - $\delta$

Percentage of Idle time = 1 - 0.42

Percentage of Idle time = 0.58

However during an 8 hour day; The amount of time that the repairman is not out on a call is = 0.58 × 8 = 4.64 hours

(d)

the probability of two or more customers in the system by assuming Poisson Distribution is:

P(N ≥ 2) = 1 - (P₀+ P₁)

where;

P₀ = 0.58

P₁ = 0.58 × 0.42 = 0.2436

P(N ≥ 2) = 1 - ( 0.58 + 0.2436)

P(N ≥ 2) = 1 - 0.8236

P(N ≥ 2) = 0.1764

Thus; the probability of two or more customers in the system is 0.1764

7 0

3 years ago