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2 years ago

An electrician worked 28 hours a week for half a year. If his hourly wage was $25, how much did he earn during this time period?

2 answers:

7 0

So, there are 52 weeks in a year, so divide that in half to get 26 total weeks that he worked. Then, multiply 26 by 28 to get the total amount of hours he worked in his time of working, which is 728 hours. Then, since his hourly wage was $25, multiply 728 by $25 to get a total of $18,200.

Fantom [35]2 years ago

3 0

Divide 365 by 2 divide that by 7 then multiply 28 and 25 then multiply your two anwsers18250

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For Joe's illness, the doctor gave him a prescription of 28 pills. He needs to take 4 pills the first day, and then 2 pills each

kiruha [24]

Answer:

12 days.

Step-by-step explanation:

I looked at it this way:

He has to take 4/28 pills the first day, so it's 28-4=24. Then I divided 24 by 2.

In simpler terms:

28-4=24

24/2

8 0

2 years ago

An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "

FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

$\large X_{sentences}$ the mean of the “sentences” group

$\large S_{sentences}$ the standard deviation of the “sentences” group

$\large X_{intentional}$ the mean of the “intentional” group

$\large S_{intentional}$ the standard deviation of the “intentional” group

Then, we can calculate by using the computer

$\large X_{sentences}=28.75$

$\large S_{sentences}=3.53553$

$\large X_{intentional}=31.625$

$\large S_{intentional}=1.40788$

$\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875$

The standard error of the difference (of the means) for a sample of size 8 is calculated with the formula

$\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}$

So, the standard error of the difference is

$\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546$

In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of 95% for the difference of the means corresponding to a level of significance of 0.05 (5%).

If this interval contains the zero, we can say there is no significant difference.

Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples.

We get the $\large t_{0.05}$ which is a value of t such that the area under the Student's t distribution outside the interval $\large [-t_{0.05}, +t_{0.05}]$ is less than 0.05.