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SOVA2 [1]
1 year ago
7

two years a go man was 6 times as old as his daughter. In 18 years, he will be twice as old as his daughter. What are their pres

ent ages
Mathematics
1 answer:
Deffense [45]1 year ago
5 0

If the man's age was 6 times as old as his daughter and will be twice after 18 years then the present age of daughter is 7 and her father is 32.

Given that age of a man was 6 times as old as his daughter and in 18 years, he will be twice as old as his daughter.

We are required to find the present ages of both daughter and her father.

Let the age of daughter before 2 years be x.

The equations are as under:

In this way the father's age was 6x.

Present age of daughter be x+2,

Present age of her father be 6x+2.

After 18 years the age of daughter be x+2+18=x+20

After 18 years the age of her father be 6x+2+18=6x+20

It is also given that the age of her father will be 2 times the age of the daughter. So, 2(x+20)=6x+20.

2x+40=6x+20

40-20=6x-2x

20=4x

x=5

Present age of daughter=5+2=7 years.

Present age of her father=6*5+2=30+2=32 years.

Hence if the man's age was 6 times as old as his daughter and will be twice after 18 years then the present age of daughter is 7 and her father is 32.

Learn more about equations at brainly.com/question/2972832

#SPJ4

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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
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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.

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