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2 years ago

Please help with this math question

Mathematics

2 answers:

Ilya [14]2 years ago

8 0

Answer:

(a) The continuously compounding account

(b) $1.39

Step-by-step explanation:

(a) At identical interest rates, continuous compounding will yield the greatest account balance when compared to periodic forms of compounding (i.e., yearly, quarterly, monthly, weekly, etc.).

(b) By compounding interest monthly, Gregory's account balance will be $6,291.61 after 10 years. By compounding interest continuously, his balance will be $6,293.00. Therefore, by choosing the continuously compounding account, he will earn $1.39 more over 10 years.

Hatshy [7]2 years ago

3 0

Answer:

(a) Account B

(b) $1.39,

Step-by-step explanation:

(a) The account with continuous compound interest will earn the most.

(b) Amount with 2.3% interest = 5000(1 + (0,023/12))^120

= $6291.61.

Amount with Continuous compound interest

= 5000e^(0.023*10) = $6293.00.

Greg earns $1.39 more with Account B.

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Answer:

$z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358$

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Step-by-step explanation:

Data given and notation

$X_{1}=253$ represent the number with no defects in sample 1

$X_{2}=196$ represent the number with no defects in sample 1

$n_{1}=300$ sample 1

$n_{2}=300$ sample 2

$p_{1}=\frac{253}{300}=0.843$ represent the proportion of number with no defects in sample 1

$p_{2}=\frac{196}{300}=0.653$ represent the proportion of number with no defects in sample 2

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

$\alpha=0.05$ significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference in the the two proportions, the system of hypothesis would be:

Null hypothesis: $p_{1} - p_2}=0$

Alternative hypothesis: $p_{1} - p_{2} \neq 0$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{253+196}{300+300}=0.748$

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358$

Statistical decision

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

$p_v =2*P(Z>5.358) = 4.2x10^{-8}$

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