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Leni [432]
2 years ago
11

What is the simplified form of 2 over x squared plus x minus 1 over x ?

Mathematics
2 answers:
sergij07 [2.7K]2 years ago
7 0

Answer:

\boxed{\boxed{\dfrac{2}{x^2+x}-\dfrac{1}{x}=\dfrac{1-x}{x(x+1)}}}

Step-by-step explanation:

The given expression is,

\dfrac{2}{x^2+x}-\dfrac{1}{x}

Factoring x^2+x,

x^2+x=x(x+1)

Hence,

=\dfrac{2}{x(x+1)}-\dfrac{1}{x}

\mathrm{Least\:Common\:Multiplier\:of\:}x\left(x+1\right),x = x\left(x+1\right)

So,

=\dfrac{2-(x+1)}{x(x+1)}

=\dfrac{2-x-1}{x(x+1)}

=\dfrac{1-x}{x(x+1)}

Dmitriy789 [7]2 years ago
4 0
I hope this helps you

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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
Please help, I’ll mark as brainliest!! The picture is above :)
Ray Of Light [21]

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable

Exact Form:

x = - 25/8 \frac{x}{y}

Decimal Form:

x = -3.125

Mixed Number Form:

x = - 3/1/8 \frac{x}{y}

Step-by-step explanation:

5 0
2 years ago
Find the average rate of
sammy [17]

Answer:

20

20, 30, 40, 50

I think this might be the answer

5 0
2 years ago
Which pair of numbers has 2 as its greatest common factor
weeeeeb [17]

Answer:

6 and 12 have the greatest common factor of 2

5 0
2 years ago
Read 2 more answers
Weather affects flights across the nation, causing delays, cancellations, and general disruption to the NAS. You have gathered a
hammer [34]

Answer:

a.) one sample t test

b.) H0 : μ = 59.3

c.) H1 : μ > 59.3

d.) μ = 59.3 ; σ = 39.84

e.) xbar = 79.4 ; s = 61.36

Test statistic = 3.16

Step-by-step explanation:

Given the sample data:

49.00 49.00 49.00 49.00 49.00 63.00 63.00 63.00 63.00 63.00 199.00 199.00 199.00 199.00 199.00 38.00 38.00 38.00 38.00 38.00 48.00 48.00 48.00 48.00 48.00 49.00 63.00 199.00 38.00 48.00

Sample size, n = 30

Using calculator :

xbar from the data above = 79.4

Standard deviation = 61.359

H0 : μ = 59.3

H1 : μ > 59.3

Test statistic :

(Xbar - μ) ÷ (σ/sqrt(n)

σ = 34.83

(79.4 - 59.3) ÷ (34.83/sqrt(30))

20.1 ÷ 6.359

Test statistic = 3.16

4 0
2 years ago
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