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The teacher select a sample of n =15 students. And he wanted to know if teaching a smaller class was more effective and students performed better, that means if $\mu >78$ and he obtaind a sample mean of $\bar X>83$ .

Based on this the correct system of hypothesis are:

Null hypothesis: $\mu =78$

Alternative hypothesis: $\mu >78$

And the best option for this case is:

a. H0: $\mu=78$ vs. H1: $\mu>78$ .

Step-by-step explanation:

For this case we have the following previous info given:

$\mu = 78$ represent the true mean

$\sigma =10$ represent the true deviation

The teacher select a sample of n =15 students. And he wanted to know if teaching a smaller class was more effective and students performed better, that means if $\mu >78$ and he obtaind a sample mean of $\bar X>83$ .

Based on this the correct system of hypothesis are:

Null hypothesis: $\mu =78$

Alternative hypothesis: $\mu >78$

And the best option for this case is:

a. H0: $\mu=78$ vs. H1: $\mu>78$ .

5 0

3 years ago

Ted invests $8,310 in a savings account with a fixed annual interest rate of 2% compounded continuously. How long will it take f

Vsevolod [243]

Answer: It will take 8 years.

Step-by-step explanation:

Equation for interest compounded continuously:

$A=Pe^{rt}$ , A = accumulated amount , P=Principal value , r =rate of interest , t= time.

Given: P= $8,310 , r = 2% , A= $9,751.88

$9751.88=8310e^{0.02t}\\\\\Rightarrow\ \dfrac{9751.88}{8310}=e^{0.02t}\\\\\Rightarrow\ 1.17351143201=e^{0.02t}$

Taking natural log on both sides

$\ln (1.17351143201)=\ln (e^{0.02t})\\\\\Rightarrow\ 0.160000478068=0.02t\\\\\Rightarrow\ t=\dfrac{0.160000478068}{0.02}\\\\\Rightarrow\ t=8.0000239034\approx8$

Hence, it will take 8 years.

8 0

3 years ago

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