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Veronika [31]
1 year ago
15

The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If

the energy stored by the spring is 12 joules, what is the approximate weight of the spring in pounds?
Mathematics
2 answers:
Lesechka [4]1 year ago
8 0

Answer:

3.12 pound.

Step-by-step explanation:

Energy due to weight, w, of the spring = 0.9 × \sqrt{E} will be represented by

W=mg=0.9\sqrt{E}

If energy stored by the spring is 12 joules.

Then weight of the spring will be

m'g = w' = 0.9\sqrt{12}

             = 0.9 × 3.46

             = 3.12 pounds

The approximate weight of the spring  is 3.12 pound.

frutty [35]1 year ago
4 0
For this case we have the following equation:
 w = 0.9* \sqrt{E}
 Where,
 w: The weight of a spring in pounds
 E: the energy stored by the spring in joules.
 Substituting values we have:
 w = 0.9* \sqrt{12}
 Making the corresponding calculation:
 w=3.12
 Answer:
 
the approximate weight of the spring in pounds is:
 
w=3.12
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Step-by-step explanation:

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Katrina has the option of an 8-year non-subsidized student loan of $32,000 at an annual interest rate of 3.5% or an 8-year subsi
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The total interest paid on the subsidized loan is USD 11,520

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Subsidized loan at 4.5% for 8years will be 4.5/100 × 32000 x 1/12= USD120/month, USD 1440/year. For 8 years, USD11,520

For non subsidized student loan at 3.5% will be 3.5/100 x 32000x 1/12= USD 93.33/month, USD 1,119per year and USD 8,959 for 8 years

6 0
1 year ago
Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. I
xxTIMURxx [149]

Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 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 is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but 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".  

Calculate the statistic

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

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

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

df=n-1=1700-1=1699  

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

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 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 is different from 634 at 10% of signficance.

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A. One day Annie weighed 24 ounces more than Benjie, and Benjie weighed 3 1/4 pounds less than Carmen. How did Annie’s and Carme
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<h2>Answer:</h2>

A.

Let Annie's weight be = a

Let Benjie's weighs = b

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a=b+24        ......(1)

Benjie weighed 3 1/4 pounds less than Carmen.  

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Now adding (1) and (2), we get

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This gives  Annie weighs 28 ounces less than Carmen.

B.

We cannot know anyone's actual weight, as we only know their relative weights.

3 0
1 year ago
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