HELPP PLSS!!!!!!! Find the volume of the following objects.

Mathematics

2 answers:

olga_2 [115]3 years ago

7 0

Answer:

First cone: 29.32

Steps:

1. 3.14(2^2)7/3

2. 3.14(4)7/3

3. 12.56(7/3)

4. 12.56(2.3

5. 29.32

Second cone: 1526.81

Divide 18 in half and so same steps as above to get the answer! :)

Have a great day!!

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bulgar [2K]3 years ago

6 0

Answer:

Cone 1 = 29.32mi || Cone 2 = 1526.81

Step-by-step explanation:

Step 1) Apply formula (volume of cone)

V= 1/3 * π * r^2 * h [OR] V= π * r^2 * h/3

Step 2) Insert numbers. (cone 1)

V= π * 2^2 * 7/3

Step 3) Insert numbers. (cone 2)

V= π * 9^2 * 18/3 (r=9 because r=1/2 base diameter of circle, 18/2=9)

Step 4) Plug and solve.

Step 5) CHECK YOUR ANSWER BEFORE SUBMISSION. Do it a few times just to make sure it checks out, and then continue!

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72 ÷(11-3²) ÷ 3 but as a fraction

Temka [501]

Let's evaluate the quantity within the parentheses. (11 - 3^2) becomes (2). Then we have:

(72 divided by 2) / 3, or 36/3, or 12.

8 0

3 years ago

Five cards are dealt from a standard 52-card deck. (a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1

labwork [276]

Answer:

Step-by-step explanation:

As there are total 52 cards in a deck and we have to draw a set of 5 cards, we can use the formula of combination to find the total number of possible ways of drawing 5 cards.

Number of ways to draw 5 cards = $N_T$

$N_T\;=\;({}^NC_k)\\\\N_T\;=\;({}^{52}C_5)\\\\N_T\;=\;2,598,960$

(a) Assuming the cards are drawn in order (would not affect the probability). The of getting Ace, 2, 3, 4 and 5 can be obtained by multiplying the probability of getting cards below 6 (20/52) with the probability of getting 5 different cards (4 choices for each card).

$P(a)\;=\;\frac{20}{52}*\frac{4}{52}*\frac{4}{51}*\frac{4}{50}*\frac{4}{49}*\frac{4}{48}\\\\P(a)\;=\; 1.3133*10^{-6}$

(b) For a straight we require our set to be in a sequence. The choices for lowest value card to produce a sequence are ace, 2, 3, 4, 5, 6, 7, 8, 9, or 10. Hence, the number of ways are $({}^{10}C_1)$ .

For each card we can draw from any of the 4 sets. It can be described mathematically as: $({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]$