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3 years ago

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If you are trying to compare 2 loans, one that pays 3.15% compounded mounthly for 4 years and another that pays 2.95% compounded

weekly for 4.5 years, what is the APY of the cheaper loan?

1 answer:

Afina-wow [57]3 years ago

3 0

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Nancy sells handcrafted bracelets at a flea market for $6. If her monthly fixed costs are $625 and each bracelet cost her $1.75

Mandarinka [93]

148, 6-1.75=5.25, then.... 625/5.25& round up.... its 148.

7 0

3 years ago

The radius of a circle is 18 miles. What is the circle's area?<br><br> Use 3.14 for (pi sign)

Lyrx [107]

Don’t click on the link, it is a virus.... the answer is 18 squared x 3.14

So then you get 324x3.14, and you get 1,017.36

That is ur answer

5 0

2 years ago

Jerry sold 7/20 of the toral number of tickets that we're sold for the spring band concert

Juli2301 [7.4K]

Question is Incomplete;Complete question is given below;

Jerry sold 7/20 of the total number of tickets that were sold for the spring band concert. What percent of the total number of tickets did jerry sell.

Answer:

Jerry sold 35 % of the total tickets.

Step-by-step explanation:

Given:

Jerry sold tickets = $\frac{7}{20}$ of total number of tickets

We need to find the percent of the total number of tickets Jerry sold.

Solution:

to find the percent of the total number of tickets Jerry sold we need to multiply the fraction by 100 we get;

framing in equation form we get;

Percent of the total number of tickets Jerry sold = $\frac{7}{20}\times100 = 35\%$

Hence Jerry sold 35 % of the total tickets.

5 0

3 years ago

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

3 years ago

Read 2 more answers

What function is shown in the graph below?

pochemuha

Answer:

C. y = log 1/6 x

Step-by-step explanation:

im literally on this right now on edge lol

6 0

2 years ago