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Luba_88 [7]
3 years ago
6

There are 25 students in Mrs. Venetozzi’s class at the beginning of the school year, and the average number of siblings for each

student is 3. A new student with 8 siblings joins the class in November.
The average number of siblings that each student has is

Select a Value
* larger than the original average.
*smaller than the original average.
*the same as the original average.
Mathematics
1 answer:
andrezito [222]3 years ago
7 0
Adding the student with eight siblings will increase the class average amount of siblings because this student has more siblings than the average.

Therefore, the average number of siblings that each student has is larger than the original average.
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Describe how you think you can use U.S. traditional multiplication to multiply a 1 digit number by any size number .
egoroff_w [7]
Well for example 4x1=4 hope this helps
8 0
3 years ago
2x - 6x + 9 factored
sergejj [24]
I might not be correct please let me know !! But as of right now I think the answer is

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7 0
3 years ago
Add 4.136+29.95 and show all your work!<br><br> *please don't go on the internet and copy and paste*
geniusboy [140]

Answer:

34.086

Step-by-step explanation:

4.136 can be written as 04.136

29.95 can be written as 29.950

In the sum digit at any position will be added at the digit at the same position

Carry \ \ \ 1\ \ \ \ 1\ \ \ \  \ \ \ \ 0\ \ \ \ 0\ \ \ \ 0\\4.136=0\ \ \ \ 4\ \ \ \ .\ \ \ \ 1\ \ \ \ 3\ \ \ \ 6\\29.95=2\ \ \ \ 9\ \ \ \ .\ \ \ \ 9\ \ \ \ 5\ \ \ \ 0\\\\=\ \ \ \ \ \ \ \ 3\ \ \ \ 4\ \ \ \ .\ \ \ \ 0\ \ \ \ 8\ \ \ \ 6

7 0
3 years ago
Consider the word grateful. if all of the letters are used, how many ways can the letters be arranged?
AlexFokin [52]

Answer:

Correct option: d -> 40,320

Step-by-step explanation:

To find all the arrangements of letters in a word where all the letters are different, we just need to calculate the factorial of the number of letters.

The word "grateful" has 8 different letters, so the number of arrangements of letters is:

8! = 8*7*6*5*4*3*2 = 40,320 different arrangements

Correct option: d

7 0
3 years ago
Let vector F = (6 x^2 y + 2 y^3 + 4 e^x) i + (7 e^{y^2} + 54 x) j . Consider the line integral of vector F around the circle of
balu736 [363]

Denote the circle of radius a by C. C is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of C (call it D):

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy

=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy

=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy

D is a circle of radius a, so we can write the double integral in polar coordinates as

\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta

a. For a=1, we have

\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}

b. Let I(a) denote the integral with unknown parameter a,

I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta

By the fundamental theorem of calculus,

I'(a)=12\pi(9a-a^3)

I(a) has critical points when

12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3

If a=0, then line integral is 0, so we ignore that critical point. For the other two, we would find I(\pm3)=243\pi.

8 0
4 years ago
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