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3 years ago

Is the ratio 25 to 16 proportional to the ratio 5 to 4

Mathematics

1 answer:

Nutka1998 [239]3 years ago

6 0

No? I'm pretty sure it wouldn't be because 25 to 16 and then 5 to 4 if you multiplied by 5 it would be 25 to 20 so no.

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A person places $4080 in an investment account earning an annual rate of 5.7%, compounded continuously. Using the formula

tangare [24]

We have been given that a person places $6340 in an investment account earning an annual rate of 8.4%, compounded continuously. We are asked to find amount of money in the account after 2 years.

We will use continuous compounding formula to solve our given problem as:

, where

A = Final amount after t years,

P = Principal initially invested,

e = base of a natural logarithm,

r = Rate of interest in decimal form.

Upon substituting our given values in above formula, we will get:

Upon rounding to nearest cent, we will get:

Therefore, an amount of $7499.82 will be in account after 2 years.

4 0

3 years ago

Read 2 more answers

What is the equation

erastova [34]

Thats figure is made of 2 triangles and 1 rectangle
so we will find the area of triangles and rectangles then add them

Triangle Area=1/2*x/2*2=x/2

multiply by 2 as we have two triangles=2*x/2=x
Area of a rectangle=x*2=2x
Add them=x+2x=14(given area of trepaziod
3x=14

hope i am correct

6 0

3 years ago

Help with math please example 1

hram777 [196]

Answer:

I'm not very sure about this one what icing in probably not going to be right but what I think is that they're not the same because look at the size difference imagine putting an adult's next to a baby that's kind of what it's like the bigger quadrilateral is also the adults while the little one is the baby what do you think about that I probably think that they're not saying

Step-by-step explanation:

sorry I gave it to you in my answer

3 0

3 years ago

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec

VMariaS [17]

Answer:

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=19.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size

Solution to the problem

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:

$df=n-1=9-1=8$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that $t_{\alpha/2}=$

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

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

5 0

3 years ago

the 41st term of the arithmetic sequence is -539 and the 54th term is -708. Find the value of the 80th term

yulyashka [42]

Answer:

$a _{n} = a _{1} + (n - 1)d \\ a _{41} = - 534 \\ a _{54} = - 708 \\ a _{41} = a _{1} + (41 - 1)d = - 534 \\ a _{1} + 40d = - 534 \: \: ...(1)\\ a _{54} = a _{1} + (54 - 1)d =- 708 \\ a _{1} + 53d = - 708 \: \: ...(2)$

Try to solve the similtanious equation to find the first term and the common difference, then use that information to figure out the 80th term... Update me if you need further help ...

7 0

3 years ago