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satela [25.4K]
3 years ago
14

You asked one of your team members about the schedule variance (SV) for one of her key deliverables. She mentioned that she is b

ehind schedule but there would not be any cost variance. Which of the following is NOT true in this case
Mathematics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

Where there is a negative schedule variance (SV), the project is said to be behind schedule

Where there is a negative cost variance (CV) the project is said to be over budget

Step-by-step explanation:

Schedule variance is found by subtracting the cost of work scheduled from the cost of performed work

It is given by;

Schedule variance (SV) = Cost of performed work - Cost of work scheduled

or

Schedule variance (SV) = BCWP - BCWS

Where

BCWP = Budgeted Cost of Work Performed

BCWS = Budgeted Cost of Work Scheduled

Schedule variance (SV) = Earned Value (EV) - Planned Value (PV)

Where there is a negative schedule variance, the project is said to be behind schedule.

Cost variance (CV) is given by;

Cost variance is found by subtracting the actual cost (AC) of a project from the expected cost (EC) of the project.

CV = EV - AC = EC - AC

Where there is a negative cost variance the project is said to be over budget.

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How many fifths are there in 3⅗​
Ilya [14]

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18/5

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5 1/5 = 1 thus 3 are 15 + 3 = 18/5

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John and Martha are contemplating having children, but John’s brother has galactosemia (an autosomal recessive disease) and Mart
Rina8888 [55]
<h2>Answer:</h2>

Probability=\frac{1}{24}

<h2>Step-by-step explanation:</h2>

As the question states,

John's brother has Galactosemia which states that his parents were both the carriers.

Therefore, the chances for the John to have the disease is = 2/3

Now,

Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.

Now, one of those children married with a person.

So,

Probability for the child to have disease will be = 1/2

Now, again the child's child (Martha) probability for having the disease is = 1/2.

Therefore,

<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>

Probability=\frac{2}{3}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{4}\\Probability=\frac{1}{24}

(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)

<em><u>Hence, the probability for the first child to have Galactosemia is \frac{1}{24}</u></em>

3 0
3 years ago
Pls help with all of this idk how to do
garik1379 [7]
4 times 1 = 4 so the top one would be 4/12, but i can't help with the rest because i don't understand what they are asking for 
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If 15 oranges cost Rs. 70,how much do 39 oranges cost ?​
Minchanka [31]

Answer:

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Step-by-step explanation:

70 ÷ 15 = 4.66

4.66 × 39 = 181.974

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