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kherson [118]
3 years ago
15

Lillian is going to invest in an account paying an interest rate of 5.7% compounded continuously. How much would Lillian need to

invest, to the nearest ten dollars, for the value of the account to reach $7,600 in 7 years?
Mathematics
1 answer:
Tems11 [23]3 years ago
5 0

Answer: $5156

Step-by-step explanation:

Given

The rate of interest is 5.7%

time t=7 years

Amount after 7 years, A=\$7600

for compound interest, the Amount is given by

\Rightarrow A=P(1+\frac{r}{100})^T

Put values

\Rightarrow 7600=P(1+0.057)^7\\\\\Rightarrow P=\dfrac{7600}{(1.057)^7}=5,155.71\\\\\Rightarrow P\approx \$5156

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A university hired you as a lab assistant after you graduated. Your starting salary is $41,500 and your salary will increase by
VladimirAG [237]

Answer:

$67,000

Step-by-step explanation:

41,000*(1,700*15)=67,000

5 0
3 years ago
Explain why the graph below does not represent a direct variation. a. the line does not intercept the x-axis c. the line does no
Rama09 [41]

The graph is not a direct variation because the line does not go through the origin; option C.

<h3>What is a direct variation?</h3>

A direct variation is a relationship between two or more quantities in which as one quantity increases, the other quantity also increases. Similarly, if the other quantity decreases, the other decreases as well.

For the graph of a direct variation, the line must pass through the origin.

Therefore, the graph does not represent a direct variation because the line does not go through the origin.

In conclusion, direct variation involves a corresponding increase or decrease in two related quantities.

Learn more about direct variation at: brainly.com/question/2633726

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3 0
2 years ago
A cylinder has a radius of 4 inches and a height of 5 inches. Which of these dimensions of a cylinder will have the same volume?
Kitty [74]
I don’t know this is a random guess but I’ll say radius = 10 inches and height= 8 inches
7 0
3 years ago
Tell which value of the variable is the solution of the equation. 6c=96; c=14,15,16,17
otez555 [7]

Answer:

c = 16

Explanation:

Do it on a calculator it'll be correct

7 0
3 years ago
A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un
WINSTONCH [101]

Answer:

Part 1: The statistic

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=35+35-2=68  

Replacing we got

t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31  

Part 2: P value  

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

p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002  

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

\bar X_{1}=5135 represent the mean for four year college

\bar X_{2}=4436 represent the mean for two year college

s_{1}=783 represent the sample standard deviation for four year college

s_{2}=553 represent the sample standard deviation two year college

n_{1}=35 sample size for the group four year college

n_{2}=35 sample size for the group two year college

\alpha=0.01 Significance level provided

t would represent the statistic (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=35+35-2=68  

Replacing we got

t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31  

Part 2: P value  

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

p_v =P(t_{68}>4.31)=0.000022  

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0
3 years ago
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