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3 years ago

If you roll two fair dice (one black die and one white die), in how many ways can you obtain a sum of at most 11?

Mathematics

2 answers:

GuDViN [60]3 years ago

7 0

There are 35 different combinations

zheka24 [161]3 years ago

6 0

There are 35 different combinations

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Francis works at Carlos Bakery and is making cookie trays. She has 48 chocolate chip cookies, 64 rainbow cookies, and 120 oatmea

amm1812

The number of cookies and trays are illustrations of greatest common factors.

The number of trays is 8
6 chocolate chips, 8 rainbows and 15 oatmeal cookies would fit each tray

The given parameters are:

$\mathbf{Chocolate\ chip=48}$

$\mathbf{Rainbow=64}$

$\mathbf{Oatmeal=120}$

(a) The number of trays

To do this, we simply calculate the greatest common factor of 48, 64 and 120

Factorize the numbers, as follows:

$\mathbf{48 = 2 \times 2 \times 2 \times 2 \times 3}$

$\mathbf{64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2}$

$\mathbf{120 = 2 \times 2 \times 2 \times 3 \times 5}$

So, the GCF is:

$\mathbf{GCF= 2 \times 2 \times 2}$

$\mathbf{GCF= 8}$

Hence, the number of tray is 8

(b) The number of each type of cookie

We have

$\mathbf{Chocolate\ chip=48}$

$\mathbf{Rainbow=64}$

$\mathbf{Oatmeal=120}$

Divide each cookie by the number of trays

So, we have:

$\mathbf{Chocolate\ chip = \frac{48}{8} = 6}$

$\mathbf{Rainbow = \frac{64}{8} = 8}$

$\mathbf{Oatmeal = \frac{150}{8} = 15}$

Hence, 6 chocolate chips, 8 rainbows and 15 oatmeal cookies would fit each tray

Read more about greatest common factors at:

brainly.com/question/11221202

4 0

2 years ago

I need to find the answer to 2/3 - 1/5. What is it?

Hunter-Best [27]

2/3 - 1/5

make a common denominator

10/15 - 3/15

simplify

7/15 is your answer

hope this helps

4 0

3 years ago

The following integral requires a preliminary step such as long division or a change of variables before using the method of par

shtirl [24]

Division yields

$\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}$

Now for partial fractions: you're looking for constants a, b, and c such that

$\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}$

$\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a$

which gives a + b = 2, c = 0, and 2a = -7, so that a = -7/2 and b = 11/2. Then

$\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}$

Now, in the integral we get

$\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx$

The first two terms are trivial to integrate. For the third, substitute y = x ² + 2 and dy = 2x dx to get

$\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}$