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50 testers can complete 25 test cases in 10 days. All testers are equally productive. A project has 25 testers and 50 test cases. How much time will they take to complete?

Answers

Answer 1

Answer:

It will take to complete cases 4 days per 5 cases.

Given that

50 testers can complete 25 test cases in 10 days.

All testers are equally productive.

A project has 25 testers and 50 test cases.

We have to determine

How much time will they take to complete?

According to the question

50 testers can complete 25 test cases in 10 days.

The number of cases completed per day is,

$= (50)/(25)\n\n= 2$

The number of cases completed per day is 2.

A project has 25 testers and 50 test cases.

The time will they take to complete is 2x.

= 2(2) = 4 days per 5 cases

Hence, they will take to complete cases 4 days per 5 cases.

To know more about Equation click the link given below.

brainly.com/question/1090830

Answer 2

Answer:

Answer:

40 Days

Step-by-step explanation:

50 testers

25 test case

-> 10 days

(2 days per 5 cases)

25 testers

50 test cases

2x the effort of 50 testers.

(4 days per 5 cases)

-> 40 days

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Find the surface area and volume of a rectangular solid with length 10, width 7, height 12surface area______square inches
volume_____cubic inches

Answers

Answer:

Surface area is =548

Volume = 840 in³

Step-by-step explanation:

l × w, l × w, w × h, w × h, l × h, l × h

70 + 70 + 84 + 84 + 120 + 120 = 548

l × w × h = 840

In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.02 margin of error and use a confidence level of 99%. Complete parts (a) through (c) below.a. Assume that nothing is known about the percentage to be estimated.n = ________b. Assume prior studies have shown that about 55% of full-time students earn bachelor's degrees in four years or less.n = _______c. Does the added knowledge in part (b) have much of an effect on the sample size?

Answers

Answer:

(a) The sample size required is 2401.

(b) The sample size required is 2377.

Step-by-step explanation:

The confidence interval for population proportion p is:

$CI=\hat p\pm z_(\alpha/2)\sqrt{(\hatp(1-\hat p))/(n)}$

The margin of error in this interval is:

$MOE=z_(\alpha/2)\sqrt{(\hatp(1-\hat p))/(n)}$

The information provided is:

MOE = 0.02

$z_(\alpha/2)=z_(0.05/2)=z_(0.025)=1.96$

(a)

Assume that the proportion value is 0.50.

Compute the value of n as follows:

$MOE=z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}\n0.02=1.96* \sqrt{(0.50(1-0.50))/(n)}\nn=(1.96^(2)*0.50(1-0.50))/(0.02^(2))\n=2401$

Thus, the sample size required is 2401.

(b)

Given that the proportion value is 0.55.

Compute the value of n as follows:

$MOE=z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}\n0.02=1.96* \sqrt{(0.55(1-0.55))/(n)}\nn=(1.96^(2)*0.55(1-0.55))/(0.02^(2))\n=2376.99\n\approx2377$

Thus, the sample size required is 2377.

(c)

On increasing the proportion value the sample size decreased.

At the start of 2014 Mike's car was worth £12000.The value of the car decreased by 30% every year.
Work out the value of his car at the start of 2017.

Answers

The value of Mike's car at the start of 2017 is £4116.

What is percentage ?

Percentage is a ratio in the form of fraction of 100.

Percentage is defined by the "%" symbol.

What is the required value of the car ?

At the start of the year 2014, Mike's car was worth £12000.

The value of the car decreased by 30% every year.

So, The value of the car at the start of 2015 = £12000× $(1-(30)/(100))$

= £ 12000× $(7)/(10)$

= £ 8400

Again, The value of the car at the start of 2016 = £8400× $(1-(30)/(100))$

= £8400× $(7)/(10)$

= £5880

∴ The value of the car at the start of 2017 = £5880× $(1-(30)/(100))$

= £5880× $(7)/(10)$

= £4116

Learn more about percentage here :

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The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

Answers

Answer:

The lifetime value needed is 11.8225 hours.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = (X - \mu)/(\sigma)$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. This means that $\mu = 11, \sigma = 1$ .