Anybody wanna c h at zoo m<br> 928-1137-7133<br> HtHq72

Nevaeh invested $660 in an account paying an interest rate of 3% compounded

iris [78.8K]

Answer:

19

Step-by-step explanation:

Rate 1: 3

4

1

%=3+1/4=

\,\,3.25\%\rightarrow 0.0325

3.25%→0.0325

\text{Rate 2: }3\tfrac{1}{8}\%=3+1/8=

Rate 2: 3

8

1

%=3+1/8=

\,\,3.125\%\rightarrow 0.03125

3.125%→0.03125

\text{Calculate Final Amount for Nevaeh}

Calculate Final Amount for Nevaeh

\overline{\phantom{\text{Calculate Final Amount for Nevaeh}}}

Calculate Final Amount for Nevaeh

\text{Compounded Continuously:}

Compounded Continuously:

A=Pe^{rt}

A=Pe

rt

P=660\hspace{35px}r=0.0325\hspace{35px}t=14

P=660r=0.0325t=14

Given values

A=660e^{0.0325(14)}

A=660e

0.0325(14)

Plug in

A=660e^{0.455}

A=660e

0.455

Multiply

A=1040.2744

Use calculator (with e button)

\text{Calculate Final Amount for Jonathan}

Calculate Final Amount for Jonathan

\overline{\phantom{\text{Calculate Final Amount for Jonathan}}}

Calculate Final Amount for Jonathan

\text{Compounded Monthly:}

Compounded Monthly:

A=P\left(1+\frac{r}{n}\right)^{nt}

A=P(1+

n

r

)

nt

Compound interest formula

P=660\hspace{35px}r=0.03125\hspace{35px}t=14\hspace{35px}n=12

P=660r=0.03125t=14n=12

Given values

A=660\left(1+\frac{0.03125}{12}\right)^{12(14)}

A=660(1+

12

0.03125

)

12(14)

Plug in values

A=660(1.0026042)^{168}

A=660(1.0026042)

168

Simplify

A=1021.6468

Use calculator

\text{How much more money Nevaeh has:}

How much more money Nevaeh has:

1040.2744-1021.6468

1040.2744−1021.6468

18.6276

\$ 19

$19

4 0

3 years ago

the 25 members of the reading club are buying a book to discuss at the next meeting but Club leader got a special price the book

Sever21 [200]

25*8.14=?

Multiply the cost of the book per person ($8.14) by the number of people (25).

6 0

3 years ago

What is 10 x 2.3-5 in standard notation?

timofeeve [1]

Answer:I think it’s option d

Step-by-step explanation:

4 0

3 years ago

Soft drink consumption in New Zealand. A survey commissioned by the Southern Cross Healthcare Group reported that 16% of New Zea

lidiya [134]

<h2>Answer with explanation:</h2>

Given : The proportion of New Zealanders consume five or more servings of soft drinks per week : $\hat{p}= 0.16$

a) The number of survey respondents reported that they consume five or more servings of soft drinks per week = 2006

b) Confidence interval for population proportion (p) :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ = Sample proportion.

n= Sample size.

z* = Critical value.

For n= 2006 , $\hat{p}= 0.16$ and critical value for 95% confidence interval : z* = 1.96

Then , the required confidence interval will be :

$0.16\pm (1.96)\sqrt{\dfrac{ 0.16(1- 0.16)}{2006}}$

$0.16\pm (1.96)\sqrt{ 0.000066999002991}$

$0.16\pm (1.96)(0.00818529186963)$

$0.16\pm 0.0160431720645$

$(0.16-0.0160431720645,\ 0.16+0.0160431720645)$

$(0.143956827935,\ 0.176043172064)\approx(0.144,\ 0.176)$

i.e. A 95% confidence interval for the proportion of New Zealanders who report that they consume five or more servings of soft drinks per week. = $(0.144,\ 0.176)$

c) $(0.144,\ 0.176)=(0.144\times100\%,\ 0.176\times100\%)$

$=(14.4\%,\ 17.6\%)$ 176\times100\%)[/tex]

d) The estimate might be biased because the survey is taken online that means offline New Zealanders are out of consideration.

6 0

3 years ago

Many people use scanners to read documents and store them into pdf file. To help determine which brand of scanner to buy, a stud

konstantin123 [22]

The data is missing in the question. The data is provided below :

Document : 1 2 3 4 5 6 7 8

Brand A 17 29 18 14 21 25 22 29

Brand B 21 38 15 19 22 30 31 37

Solution :

State of the hypothesis of the null hypothesis and alternate hypothesis.

Null hypothesis : $h_A = h_B$

Alternate hypothesis : $h_A > h_B$

These hypothesis is a one tailed test. The null hypothesis will get rejected when the mean difference between the sample means is very small.

Significance level = 0.05

Therefore the standard error is : $SE = \sqrt{(\frac{s^2_1}{n_1})+(\frac{s^2_2}{n_2})}$

= 3.602

And the degree of freedom, DF = 14

$t=\frac{(x_1-x_2)-d}{SE}$

= -1.319

Here, $s_1$ = standard deviation of the sample 1

$s_2$ = standard deviation of the sample 2

$n_1$ = size of the sample 1

$n_2$ = size of the sample 2

$x_1$ = mean of the sample 1

$x_2$ = mean of the sample 2

d = the hypothesis difference between the population mean

The observed difference in a sample means t static of -1.32. From t distribution calculator to determine P( $t \leq -1.32$ ) = 0.1042

Since the P value of 0.1042 is greater than significance level o 0.05, we therefore cannot reject the null hypothesis.

But from the test, we have no sufficient evidence that supports that Brand A is better than Brand B.

8 0

3 years ago

Anybody wanna c h at zoo m 928-1137-7133 HtHq72