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S_A_V [24]
3 years ago
9

After 1 year, $600 deposited in a savings account with simple interest had earned $84 in interest. What was the interest rate?​

Mathematics
1 answer:
EleoNora [17]3 years ago
6 0

Answer:

14%

Step-by-step explanation:

For this question you need to find out what percent 84 is of 600,

If ony $100 was deposited, then $84 would equal 84%.

If $200 was deposited, then $84 would be 42% which is half of 84, and so on...

So $84 is 14% of $600 becuase 84 divided by 6 is 14.

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Answer: $526.39

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4(4n-3)=24+4n I'm not sure how to solve this
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Answer: n=3

Step-by-step explanation:

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You are paid $119 for 8.50 hours of work. what is your rate of pay?
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The​ quality-control manager at a compact fluorescent light bulb​ (CFL) factory needs to determine whether the mean life of a la
Illusion [34]

Answer:

a) Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 7539

b) t=\frac{7359-7539}{\frac{840}{\sqrt{49}}}=-1.5    

The degrees of freedom are given by:

df=n-1=49-1=48  

The p value is given by:

p_v =P(t_{(48)}  

c) 7359-2.01\frac{840}{\sqrt{49}}=7117.8    

7359+2.01\frac{840}{\sqrt{49}}=7600.2    

d) For this case since the confidence interval contains the value of 7539 we don't have enough evidence to reject the null hypothesis at the significance level given of 5%. same conclusion using the hypothesis test and with the confidence interval

Step-by-step explanation:

Part a and b

Data given

\bar X=7359 represent the sample mean

\sigma=840 represent the population standard deviation

n=49 sample size  

\mu_o =7539 represent the value to test

t would represent the statistic  

p_v represent the p value

Hypothesis to verify

We want to verify if the mean life is different from 7539 hours, the system of hypothesis would be:  

Null hypothesis:\mu \geq 7539  

Alternative hypothesis:\mu < 7539  

The statistic for this case would be given by:

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

Replacing the info given we got:

t=\frac{7359-7539}{\frac{840}{\sqrt{49}}}=-1.5    

The degrees of freedom are given by:

df=n-1=49-1=48  

The p value is given by:

p_v =P(t_{(48)}  

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 7539

Part c

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

We can find the critical value using the confidence level given of 95% and using the t distribution with 48 degrees of freedom we got t_{\alpha/2}=\pm 2.01

Now we have everything in order to replace into formula (1):

7359-2.01\frac{840}{\sqrt{49}}=7117.8    

7359+2.01\frac{840}{\sqrt{49}}=7600.2    

Part d

For this case since the confidence interval contains the value of 7539 we don't have enough evidence to reject the null hypothesis at the significance level given of 5%. same conclusion using the hypothesis test and with the confidence interval

3 0
3 years ago
Sammy was selling tickets for the football game. He sold five more adult tickets than senior tickets and twice as many children'
tester [92]

Sammy was selling tickets for the football game. He sold five more adult tickets (AT) than senior tickets (ST) and twice as many children's tickets (CT) than senior tickets. Let s represent the number of senior tickets sold

ST = AT - 5

CT = 2ST = 2(AT - 5)

4 0
4 years ago
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