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SCORPION-xisa [38]

3 years ago

12

(NO FILES OR LINKS OR RANDOM WORDS)

1 answer:

ehidna [41]3 years ago

3 0

Answer:

Randomly selecting a six of diamonds - 1 / 52

Randomly selecting a 7, 8, 9 or 10 - 4 / 13

Step-by-step explanation:

There is only 1 six of diamonds in a standard deck of cards. There are 52 cards in a deck, thus the probability of pulling a six of diamonds is 1 in 52.

There are 4 of each card in a deck. so they are 4 7's, 4 8's. 4 9's and 4 10's. And there are a total of 52 cards in a deck. So the probability of pulling a 7,8,9 or 10 are 4 + 4 + 4 + 4 in 52

4 + 4 + 4 + 4 = 16

16 / 52 simplified is 4 / 13 Therefore the is a 4 in 13 chance of pulling a 7 8 9 or 10

The other ones are correct

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Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

Order the numbers from least to greatest.<br> 23% , 1/4 , 0.225 .

Alla [95]

0.225, 23%, 1/4

Step my step

23% = 0.23
1/4 = 0.25

3 0

2 years ago

A college basketball game is 40 minutes long. Derrick’s favorite player participated for 60% of last week’s game. How many minut

alexandr1967 [171]

Answer:

24 minutes

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A company has 200 machines. Each machine has 12% probability of not working.

shutvik [7]

Answer:0.18665

2.0.00602

3.

1.0000

Step-by-step explanation:

8 0

3 years ago

Convert 1.835 × 10−4 to standard notation.

Law Incorporation [45]

<span>0.0001835 is the right answer after converting</span>

7 0

3 years ago

Read 2 more answers