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2 years ago

13

There are two options for borrowing $5,000 from a bank: Bank A = 3% simple interest for 5 years Bank B = 2.5% simple interest fo

r 6 years Which is a better deal?
They are the same
Bank B
Bank A

2 answers:

motikmotik2 years ago

8 0

Answer:

the same

Step-by-step explanation:

3*5=15

2.5*6=15

Andrew [12]2 years ago

6 0

BANK A:
5,000*0.03*5=750
BANK B:
5,000*0.025*6=750

They are the same

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A car starts with a dull tank of gas. After driving around 5he city, 1/7 of the gas has been used. With the rest of the gas in t

Sati [7]

Answer:

$\frac{2}{7}$

Step-by-step explanation:

Given:

A car starts with a dull tank of gas

1/7 of the gas has been used around the city.

With the rest of the gas in the car, the car can travel to and from Ottawa three times.

Question asked:

What fractions of a tank of gas does each complete trip to Ottawa use?

Solution:

Fuel used around the city = $\frac{1}{7}$

Remaining fuel after driving around the city = 1 - $\frac{1}{7}$

= $\frac{7 - 1}{7} = \frac{6}{7}$

According to question:

As from the rest of the gas in the car that is $\frac{6}{7}$ , the car can complete 3 trip to Ottawa which means,

By unitary method:

The car can complete 3 trip by using = $\frac{6}{7}$ tank of gas.

The car can complete 1 trip by using = $\frac{6}{7} \div 3$

= $\frac{6}{7} \times\frac{1}{3}$

= $\frac{6}{21}$

= $\frac{2}{7}$ tank of gas

Thus, $\frac{2}{7}$ tank of gas used for each complete trip to Ottawa.

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3 years ago

What is the volume of the square pyramid with base edges 12cm height 10cm

Alina [70]

V=480cm³ . 10 and 12

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3 years ago

Use the fundamental counting principle to determine all the possible outcomes from rolling a die once then spinning the spinner.

Rudik [331]

The answer is there are 6 possible outcomes: {1, 2, 3, 4, 5, 6} Hope this helps!

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2 years ago

An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co

levacccp [35]

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)

2. $4.7048 \leq \sigma^2 \leq 16.1961$

3. $t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

$t_{crit}=1.753$

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. $t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

$\chi^2 =24.9958$

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

$\bar X=26.31$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=16-1=15$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that $t_{\alpha/2}=2.13$

Now we have everything in order to replace into formula (1):

$26.31-2.13\frac{2.8}{\sqrt{16}}=24.819$

$26.31+2.13\frac{2.8}{\sqrt{16}}=27.801$

So on this case the 95% confidence interval would be given by (24.8190;27.8010)

Part 2

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

$\chi^2_{\alpha/2}=24.996$

$\chi^2_{1- \alpha/2}=7.261$

And replacing into the formula for the interval we got:

$\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}$

$4.7048 \leq \sigma^2 \leq 16.1961$

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:

Null hypothesis: $\mu \leq 25$

Alternative hypothesis: $\mu > 25$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

Critical value

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got $t_{crit}=1.753$

Conclusion

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: $\sigma \leq 2.3$

H1: $\sigma >2.3$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be $\chi^2 =24.9958$

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

7 0

3 years ago

3 points lie on an xy plane, with each point representing where a particular person lives. Amy lives at point (2,3); Brian lives

geniusboy [140]

The equidistant point for all three, will be the "centroid" of their triangular locations
thus $\bf \textit{Centroid of a Triangle}
\begin{array}{llll}
Amy
\begin{array}{llll}
(2,3)\\
x_1,y_1
\end{array}\quad Brian
\begin{array}{llll}
(0,7)\\
x_2,y_2
\end{array}\quad Claire
\begin{array}{llll}
(-2,3)\\
x_3,y_3
\end{array}\\ \quad \\\\\\
\left(\cfrac{x_1+x_2+x_3}{3}\quad ,\cfrac{y_1+y_2+y_3}{3}\quad \right)
\end{array} 
$

6 0

2 years ago