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Dimas [21]
4 years ago
7

Will earned 750 per week. If he earns 20. per hour how many hours did he work

Mathematics
1 answer:
Alenkinab [10]4 years ago
5 0

Answer:

He worked 37.5 hours.

Step-by-step explanation:

Step 1- Divide.

750÷20= 37.5

So, he worked 37 and a half hours.

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Someone please help I really appreciate it
Alex_Xolod [135]
So you have to find the relationship between the pumpkin diameter (y) and the number of weeks passed (x) in a table, lets do it taking into account that the equation modeling such behaviour is:
y = 2x + 6, where x is the number of weeks and plus original 6 cm
x                                y                            
week                         diameter
0                                6
1                                8
3                                12
5                                16
10                               26
substitute the x and y values in the equation to see how they fit into it
5 0
3 years ago
Read 2 more answers
While researching the cost of school lunches per week across the state, you use a sample size of 45 weekly lunch prices. The sta
Drupady [299]

We assume the lunch prices we observe are drawn from a normal distribution with true mean \mu and standard deviation 0.68 in dollars.


We average n=45 samples to get \bar{x}.


The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write


\sigma = 0.68 / \sqrt{45} = 0.101


Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains \mu.


Our interval takes the form of ( \bar{x} - z \sigma, \bar{x} + z \sigma ) as \bar{x} is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".


Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.


Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.


With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.


We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is


( \bar{x} - 1.65 (.101),  \bar{x} + 1.65 (.101) )


in other words a margin of error of


\pm 1.65(.101) = \pm 0.167 dollars


That's around plus or minus 17 cents.




3 0
3 years ago
Read 2 more answers
The yearbook club had a meeting. the meeting had 18 people, which is one half of the club. how many people are in the club
Dmitry_Shevchenko [17]

Answer:

36 people

Step-by-step explanation:

If the yearbook club had 1/2 the participants attend the meeting, than all of them would be represented by 18 x 2, or 36 (since two halves make a whole)

8 0
2 years ago
2 less then 5 times a number
Alex

Answer:

5-2n

Step-by-step explanation:

4 0
4 years ago
Look at the equation below. Identify which equation has no solution
IgorC [24]

Answer:

C

Step-by-step explanation:

24a-22  = -4(1-6a)\\\\\implies 24a-22  = -4+24a\\\\\implies -22 = -4\\\\\text{Which is not true, so no solutions exist.}

4 0
2 years ago
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