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Mathematics

1 answer:

krek1111 [17]3 years ago

7 0

Answer:

3 : 2

divide both by 15

Step-by-step explanation:

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A gardener transplanted a plant from an indoor pot to an outdoor garden. After 30 days, the plant's height was 18 centimeters. A

Aliun [14]

Let x is the days and the y is the height of the plant.

Let the change between x and y is linear.

So, the relation between x and y will take the form ⇒ y = ax + b

where a and b are constants.

At x = 30 ⇒ y = 18 ⇒⇒⇒ ∴ 18 = 30 a + b → (1)

At x = 90 ⇒ y = 29 ⇒⇒⇒ ∴ 29 = 90 a + b → (2)

solve (1) and (2) to find a and b

subtract (1) from (2)

∴ 11 = 60 a ⇒⇒⇒ ∴ a = 11/60

substitute at (1) ⇒⇒⇒ ∴ b = 25/2 = 12.5

<u>So, the equation which represents the plants growth is</u>

4 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

Rudik [331]

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: $({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]$