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PIT_PIT [208]
3 years ago
6

Eric’s mother wants to help him with his math homework. She puts 24 cookies in the cookie jar. 12 of the cookies are chocolate c

hip, 8 are oatmeal and 4 are peanut butter. She then has Eric select a cookie from the jar without looking. Next, without replacing the first cookie, Eric picks a second cookie without looking in the jar. What is the probability that Eric will pick an oatmeal cookie first and a chocolate chip cookies second?
Mathematics
2 answers:
Aloiza [94]3 years ago
4 0

Answer:

4/23

Step-by-step explanation:

there are 8 oatmeal cookies out of the 24 in the jar so the probability of pulling an oatmeal cookie out is 8/24 = 1/3

Then we have 23 cookies in the jar and 12 are chocolate chip

1/3 * 12/23 = 4/23

shusha [124]3 years ago
3 0

Answer:

12% 4% 8%

Step-by-step explanation:

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Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
Seven students chose integers: -16, 12, -13, -6, 5, 6 and 1 Order the numbers from least to greatest. When all numbers are order
Lorico [155]

Answer:-16<-13<-6<1<5<6<12

1

-16

12

Step-by-step explanation:

8 0
3 years ago
write a problem that can be solved using a line plot. Draw and label the line plot and solve the problem.
Debora [2.8K]
0/5 1/5 2/5 3/5 4/5 5/5  hope this helps

5 0
3 years ago
PLEASE HELP MEEEEEEEE ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Archy [21]

Answer:

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6 0
2 years ago
Read 2 more answers
Math related!!!!! Pls help look at pic &gt;&gt;&gt;&gt;&gt;
olga_2 [115]
Hey! I’m not sure what you’re needing help on but if it’s an answer i’m pretty sure the answer would be - 180 (hx)
5 0
2 years ago
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