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vovikov84 [41]
3 years ago
8

Whats the value of 4x²-3 when x=5, x=10

Mathematics
2 answers:
slava [35]3 years ago
4 0
4(5) to the power of 2 -3
:20 to the power of 2 =400
:400-3=397

4(10) to the power of 2 -3
:40 to the power of 2=1600
:1600-3=1597
Pepsi [2]3 years ago
3 0
Just replace 5 and 10 in the equation
x = 5, 4(5)^2-3 = 97
x = 10, 4(10)^2-3 = 397
Hope it helped
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Step-by-step explanation:

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If varies inversely as . It is known that = 10 when = 3, find the value of when = 5.
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Answer:

y = 6

Step-by-step explanation:

If x varies inversely proportional as y.

x=\dfrac{k}{y}, k is constant of proportionality

or

k = xy

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A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len
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Answer:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

df=n-1=93-1=92  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

\bar X=5.97 represent the sample mean for the length

s=0.4 represent the sample standard deviation

n=93 sample size  

\mu_o =6 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:  

Null hypothesis:\mu = 6  

Alternative hypothesis:\mu \neq 6  

since we don't know the population deviation the statistic is:

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

Replacing in formula (1) we got:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

P value

The degrees of freedom are given by:

df=n-1=93-1=92  

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

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

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