Brad wants to buy flowers for

Help Please !!

Kazeer [188]

Answer:

10

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

There are 12 marbles in a box. 8 are purple, and 4 are green. What is the probability of picking a purple marble and then a gree

Bad White [126]

Answer: 8/33

Step-by-step explanation:

So the probability of picking a purple marble is 8/12, after that there are 11 marbles left in the box

Now the probability of picking a green marble without placing the first marble back is 4/11

So the probability of doing both is 8/12 x 4/11 = 8/33

8 0

2 years ago

Helppp with this question plzz

Gekata [30.6K]

Answer:

306

Step-by-step explanation:

There are 102 triangles, a triangle has 3 sides, each side is 1 cm. 102 x 3=306

6 0

3 years ago

A coin is tossed and a single 6 sided number cube is rolled. Find the probability of landing on the heads side of the coin and r

Delicious77 [7]

Answer:

The probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

Step-by-step explanation:

Given:

Coin is tossed and a single 6 sided number cube is rolled.

Let the event of tossing head be 'H' and event of rolling a 3 be 'R3'.

Now, we know that:

Probability of an event 'A' = Favorable outcomes of 'A' ÷ Total possible outcomes

Now, for tossing a coin, the total possible outcomes are head and tail. So, there are 2 possible outcomes.

So, probability of event 'H' is given as:

$P(H)=\dfrac{n(H)}{n(S)}\\\\P(H)=\frac{1}{2}=0.5$

Similarly, for rolling a 6 sided cube, the possible outcomes are numbers 1 to 6. So, the number of possible outcomes is, $n(S)=6$

Now, probability of rolling a 3 is given as:

$P(R3)=\frac{n(R3)}{n(S)}\\\\P(R3)=\frac{1}{6}$

Now, both the events 'H' and 'R3' occur together. So, the combined probability is the product of two individual probabilities as they are independent events. So,

$P(H\ and\ R3)=P(H)\times P(R3)\\\\P(H\ and\ R3)=\frac{1}{2}\times \frac{1}{6}\\\\P(H\ and\ R3)=\frac{1}{12}=0.083$

Note: Independent events are those events whose intersection is an empty set of events or the outcome of one event doesn't affect the outcome of the other event.

Therefore, the probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

4 0

3 years ago

A pyramid-shaped vat has square cross-section and stands on its tip. The dimensions at the top are 2 m × 2 m, and the depth is 5

Strike441 [17]

Answer:

the water level increases at a rate of 1.718 m/min when the depth is 4 m

Step-by-step explanation:

the volume of the pyramid is

V = (1/3)(height)(area of base) = 1/3 H*L²

For the diagonal in the pyramid

tg Ф = Side Length/ Height = L / H = x / h

where h= depth of water , x= side of the corresponding cross section

therefore x= L *h/H

the volume of the water is

v= 1/3 h*x² = 1/3 (L/H)² h³

in terms of time

v = Q*t

then

Q*t = 1/3 (L/H)² h³

h³ = 3*(H/L)² *Q *t

h = ∛(3*(H/L)² *Q *t) = ∛(3*(H/L)² *Q) *∛t = k* ∛t , where k=∛(3*(H/L)² *Q)

h = k* ∛t

then the rate of increase in depth is dh/dt

dh/dt = 1/3*k* t^(-2/3)

since

t = (h/k)³

dh/dt = 1/3*k* t^(-2/3) = 1/3*k* (h/k)³ ^(-2/3) = 1/3*k* (h/k)^(-2) = 1/3 k³ / h²

= 1/3 (3*(H/L)²*Q) / h² = (H/L)²*Q /h²

dh/dt= [H/(h*L)]²*Q

replacing values, when h=4m

dh/dt= [H/(h*L)]²*Q = [5m/(4m*2m)]² * (3m³/min)= 1.718 m/min

8 0

3 years ago