All categories

3 years ago

111. Dani has $1,000 in an investment account that earns 3% per year, compounded monthly.

Mathematics

1 answer:

tangare [24]3 years ago

5 0

Answer:

$A(t) = 1000(1.0025)^{12t}$

Step-by-step explanation:

The compound interest formula is given by:

$A(t) = P(1 + \frac{r}{n})^{nt}$

Where A(t) is the amount of money in the account after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the number of years the money is invested or borrowed for.

For this problem, we have that:

$P = 1000$

The investment is compounded monthly. There are 12 months in a year. So $n = 12$

The interest rate is 3%. So $r = 0.03$ .

The amount of money in her account after t years is:

$A(t) = P(1 + \frac{r}{n})^{nt}$

$A(t) = 1000(1 + \frac{0.03}{12})^{12t}$

$A(t) = 1000(1.0025)^{12t}$

You might be interested in

(-3 squared +3) divided by (10 – 6)

leonid [27]

Answer:

research it many my answer will be wrong

4 0

2 years ago

Read 2 more answers

3. A jet traveled at an average speed of 681 kilometers an hour. At that rate, how far did the jet go in 7 1/4 hours?

Burka [1]

$\stackrel{mixed}{7\frac{1}{4}}\implies \cfrac{7\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{29}{4}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{ccll} K ms&hours\\ \cline{1-2} 681 & 1\\ x& \frac{29}{4} \end{array} \implies \cfrac{681}{x}=\cfrac{1}{~~ \frac{29}{4}~~}\implies \cfrac{684}{x}=\cfrac{4}{29} \\\\\\ 19836=4x\implies \cfrac{19836}{4}=x\implies 4959=x$

6 0

2 years ago

A study of class attendance and grades among first-year students at a state university showed that, in general, students who mis

ratelena [41]

Answer:

the numerical value of the correlation between percent of classes attended and grade index is r = 0.4

Step-by-step explanation:

Given the data in the question;

we know that;

the coefficient of determination is r²

while the correlation coefficient is defined as r = √(r²)

The coefficient of determination tells us the percentage of the variation in y by the corresponding variation in x.

Now, given that class attendance explained 16% of the variation in grade index among the students.

coefficient of determination is r² = 16%

The correlation coefficient between percent of classes attended and grade index will be;

r = √(r²)

r = √( 16% )

r = √( 0.16 )

r = 0.4

Therefore, the numerical value of the correlation between percent of classes attended and grade index is r = 0.4

3 0

3 years ago

Women athletes at the a certain university have a long-term graduation rate of 67%. Over the past several years, a random sample

lana [24]

Answer:

$z=\frac{0.579 -0.67}{\sqrt{\frac{0.67(1-0.67)}{38}}}=-1.193$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.1$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the proportion of women athletes graduated is not significantly lower than 0.67 or 67% at 10% of significance

Step-by-step explanation:

Data given and notation

n=38 represent the random sample taken

X=22 represent the number of women athletes graduated

$\hat p=\frac{22}{38}=0.579$ estimated proportion of women athletes graduated

$p_o=0.67$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is lower than 0.67 or no:

Null hypothesis: $p \geq 0.67$

Alternative hypothesis: $p < 0.67$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.579 -0.67}{\sqrt{\frac{0.67(1-0.67)}{38}}}=-1.193$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.1$ . The next step would be calculate the p value for this test.

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

$p_v =P(z$

3 0

3 years ago

Avanety of two types of snack packs are delivered to a store. The box plots compare the number of calories in each

AveGali [126]

Answer:

The number of calories in the packs of trail mix have a greater variation than the number of calories in the packs

of crackers

Step-by-step explanation:

IQR of trail mix data = 105 - 90 = 15

The range of cracker data = 100 - 70 = 30.

Therefore, the first option is NOT TRUE.

To check if option 2 is correct, calculate the lower limit to see if 70 is below the lower limit. If 70 is below the lower limit, then it is an outlier in the trail mix data.

Thus, Lower Limit = $Q_1 - 1.5(IQR)$

Q1 = 90,

IQR = 105 - 90 = 15

Lower Limit = $90 - 1.5(15)$

Lower Limit = $90 - 22.5 = 67.5$

70 is not less than the lower limit, therefore, 70 is not an outlier for the trail mix data. The second option is NOT TRUE.

The upper quartile of the trail mix data = 105.

The maximum value of the cracker data = 100.

Therefore, the third option is NOT TRUE.

Range can be used to determine how much variable there is in a data represented on a box plot. The greater the range value, the greater the variation.

Range of trail mix data = 115 - 70 = 45

Range of cracker data = 100 - 70 = 30.

The range value for the number of calories in trail mix is greater than that for cracker, therefore, the number of calories in the packs of trail mix have a greater variation than the number of calories in the packs

of crackers.

The fourth option is TRUE.

7 0

3 years ago

Read 2 more answers