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denis23 [38]
3 years ago
5

A parcel delivery company delivered 103,000 packages last year, when its average employment was 84 drivers. This year the firm h

andled 112,000 deliveries with 96 drivers. What was the percentage change in productivity over the past two years
Mathematics
1 answer:
Aleksandr [31]3 years ago
8 0

Answer: 4.85%

Step-by-step explanation: productivity = parcel delivered / number of drivers.

At when parcel delivered was 103, 000 the average number of drivers was 84 thus making the productivity to be

103,000/ 84 = 1226.19

At when the parcel delivered was 112, 800 the average number of drivers was 96 thus making the productivity to be

112, 000/ 96 = 1166.67

Thus the change in productivity = 1226.19 - 1166.67 = 59.52.

Initial productivity = 1226.19

final productivity = 1166.67

% change in productivity = change in productivity / initial productivity * 100

% change in productivity = 59.52 /1226 * 100

% change in productivity = 4.85%

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Answer:

Explained below.

Step-by-step explanation:

According to the Central Limit Theorem if an unknown population is selected with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from this population with replacement, then the distribution of the sample means will be approximately normal.

Then, the mean of the sampling distribution of sample means is given by,

\mu_{\bar x}=\mu

And the standard deviation of the sampling distribution of sample means is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

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Can anyone help me? no links
NNADVOKAT [17]

Answer:

Step-by-step explanation:

Sum: addition

Difference: subtraction

Product: multiplication

Number                    Sum                    Difference                    Product

1, 4                         1 + 4 = 5                     4 - 1 = 3                       1 * 4 = 4

2, 8                       2 + 8 = 10                   8 - 2 = 6                      2 * 8 = 16

Hope this helps!

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Suppose you take an SRS of size 1000 of Connecticut students eligible to take the SAT and find that 85% plan to take the SAT dur
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Answer:

\mu_{p}=p=0.81

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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So under the null hypothesis the mean for the population proportion is p

\mu_{p}=p=0.81

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\sigma_{p}=\sqrt{\frac{p_0(1-p_o)}{n}}=\sqrt{\frac{0.81(1-0.81)}{1000}}=0.0124

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