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netineya [11]
2 years ago
14

High of the following described the role of mathematics in science

Mathematics
1 answer:
SpyIntel [72]2 years ago
5 0
Mathematics and science go hand in hand. The branches of science which include: Physics, astronomy, chemistry, biology, etc. In all these fields you will find that you will have to use the mathematical principles and formulae to solve the numerical problems. So, science needs maths but maths does not need science.
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The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school o
timurjin [86]

Answer:

a) Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

b) p_v =2*P(t_{(24)}  

c) If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

Step-by-step explanation:

1) Data given and notation  

77600 ,76000 ,90700 ,97200 ,90700 ,101800 ,78700 ,81300 ,84200 ,97600 ,

77500 ,75700 ,89400 ,84300 ,78700 ,84600 ,87700 ,103400 ,83800 ,101300

94700 ,69200 ,95400 ,61500 ,68800

We can calculate the sample mean and deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

The values obtained are:

\bar X=85272 represent the mean annual salary for the sample  

s=11039.23 represent the sample standard deviation for the sample  

n=25 sample size  

\mu_o =90000 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean salary differs from 90000, the system of hypothesis would be:  

Null hypothesis:\mu = 90000  

Alternative hypothesis:\mu \neq 90000  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{85272-90000}{\frac{11039.23}{\sqrt{25}}}=-2.141    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(24)}  

Part c: Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

6 0
3 years ago
Can you please tell me how to solve this?
Sedaia [141]

The answer is 142°


Explanation:

angle 1 and angle 2 form are supplementary.

m∠1 + m∠2 = 180°

38° + m∠2 = 180°

m∠2 = 180° - 38°

m∠2 = 142°


Hope it helps!

5 0
2 years ago
I need to do these factor trees for 60 and 96
Arisa [49]
60
2 and 30
30
2 and 15
15
3 and 5
96
8 and 12
circle for 8 is 2 and the square is 4
circle for 12 is 2 and the last circle is also 2
7 0
2 years ago
Smooth transitions in an informative essay_____
WITCHER [35]
The answer is connect.
hope your day is good !!
3 0
2 years ago
Read 2 more answers
Show that in any group of 101 people there are at least two persons having the same number friends (It is assumed that if a pers
vladimir1956 [14]

Answer:

Step-by-step explanation:

Let us assume that if possible in the group of 101, each had a different number of friends.  Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.

Say P has 100 friends then P has all other persons as friends.  In this case, there cannot be any one who has 0 friend.  So a contradiction.  Hence proved

Part ii: EVen if instead of 101, say n people are there same proof follows as

if different number of friends then they would be 0,1,2...n-1

If one person has n-1 friends then there cannot be any one who does not have any friend.

Thus same proof follows.

7 0
3 years ago
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