50 people sponsor paula and she raises a total of £180 for charity. what is the mean average amount sponsored?

Mathematics

1 answer:

Sveta_85 [38]3 years ago

7 0

180 / 50= £ 3.60 per person

You might be interested in

Alan has to finish an exam within 120 minutes. He finished the exam in 80 minutes. What is the percentage of the time he took fo

drek231 [11]

Answer:

66.67%

Step-by-step explanation:

80/120 = 66.67

7 0

3 years ago

In mid-day meal scheme 1/ 10 litre of milk is given to each student of a primary school. If 30 litres of milk are distributed ev

nirvana33 [79]

Answer:

300 students

Step-by-step explanation:

To know the number of students, we must divide the total amount of milk spent by the amount spent for each and in this way we will obtain the value we are looking for, like this (both have the same units of L, so there is no problem ):

30 / (1/10) = 300

that is to say that there are a total of 300 students

5 0

3 years ago

Which conversion on the metric table need division

vladimir1956 [14]

If you are going from a larger unit to a smaller unit, you multiply. If you are going from a smaller unit, you divide

5 0

3 years ago

A chemist examines 15 geological samples for potassium chloride concentration. The mean potassium chloride concentration for the

NikAS [45]

Answer:

$0.376-2.145\frac{0.0012}{\sqrt{15}}=0.375$

$0.376+2.145\frac{0.0012}{\sqrt{15}}=0.377$

And we are 95% confident that the true mean of chloride concentration is between 0.375 and 0.377 cc/ cubic meter

Step-by-step explanation:

Data provided

$\bar X=0.376$ represent the sample mean for the chloride concentration

$\mu$ population mean (variable of interest)

s=0.0012 represent the sample standard deviation

n=15 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since we need to find the critical value first we need to begin finding the degreed of freedom

$df=n-1=15-1=14$

The Confidence is 0.95 or 95%, the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ , and the critical value for this case with a t distribution with 14 degrees of freedom is $t_{\alpha/2}=2.145$