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Nana76 [90]
3 years ago
8

A $1500 loan has an annual interest rate of

Mathematics
1 answer:
Goshia [24]3 years ago
4 0

127.50 = 1,500(0.0425) * x

127.50 = 63.75 * x

x= 127.50/63.75

<span>x=2 </span>

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Polynomials <br>PLZ QUICK 1: Subtract and Simplify <br><br>(–y^2 – 4y – 8) – (–4y^2 – 6y + 3)
frez [133]
-y^2-4y-8 +4y^2+6y-3
=3y^2+2y-11
Hope this helps
7 0
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For every 4 of my wonderful nieces, I want to buy 5 Friendly cupcakes. And I have 12 wonderful nieces.
almond37 [142]

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15 cupcakes

Step-by-step explanation:

I'm guessing you mean that you want to find out how many cupcakes you'll need to buy.

4 0
3 years ago
A market research analyst claims that 32% of the people who visit the mall actually make a purchase. You think that less than 32
denis23 [38]

Answer:

The null hypothesis is H_0: p \geq 0.32

The alternative hypothesis is H_1: p < 0.32

Step-by-step explanation:

State the null and alternative hypotheses to test that less than 32% of mall visitors make a purchase

At the null hypothesis, we test if the proportion is of at least 32%, that is:

H_0: p \geq 0.32

We are testing if less than 32% of visitors make a purchase, so, at the alternative hypothesis, we test if the proportion is less than 32%, that is:

H_1: p < 0.32

5 0
3 years ago
A 12 ounce ice cream sundae is 3/4 ice cream. How many ounces of ice cream does the sundae contain?
andreyandreev [35.5K]

Answer:

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5 0
3 years ago
Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 2yi + xzj + (x
Lady_Fox [76]

By Stokes' theorem, the line integral of \vec F over C is given by the surface integral of the curl of \vec F over S, where S is the region of intersection of the plane z=y+6 and the cylinder x^2+y^2=1 with S having positive/upward orientation.

Parameterize S by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+6)\,\vec k

with 0\le u\le1 and 0\le v\le2\pi.

Take the normal vector to S to be

\vec s_u\times\vec s_v=-u\,\vec\jmath+u\,\vec k

The curl of \vec F is

\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-2)\,\vec k

Then the line integral is equivalent to

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{2\pi}\int_0^1\bigg((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+4)\,\vec k\bigg)\cdot\bigg(-u\,\vec\jmath+u\,\vec k\bigg)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^1(5u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{5\pi}

4 0
3 years ago
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