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2 years ago

Mohamed plants 30 roses in 5 plots,if he needs to plant 120 roses,how many plots he needs ?

Mathematics

1 answer:

Margarita [4]2 years ago

5 0

$[Hello,BrainlyUser]$

Answer:

20 Plots

Step-by-step explanation:

Given:

Plants 30 roses in 5 plots

Plants 120 roses in [?] Plots

Question:

How many plots he needs ?

Solve:

$\frac{roses}{plots}=Solution$

Divide 30 roses by 5 plots = 6

Hence, 120 roses divide by 6

$\frac{120}{6}=20$

Therefore, Mohamed need 20 Plots.

$[CloudBreeze]$

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Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

4 years ago

Hint: You won't flip the sign in every problem. ONLY FUP it when you or by a 9 - 2x < 6x - 15 Solutions You solve and graph t

svetoff [14.1K]

You only flip the inequality sign when you multiply or divide both sides by a negative number.

================================================

Problem 1

$3 - 8x \ge -29\\\\-8x \ge -29-3\\\\-8x \ge -32\\\\x \le \frac{-32}{-8} \ \text{ inequality sign flip}\\\\x \le 4$

The inequality sign flip happens because we divided both sides by -8.

The graph will have a closed circle at 4 with shading to the left.

Three solutions are x = 0, x = 1, x = 2. You can pick any three numbers you want as long as they are 4 or smaller.

================================================

Problem 2

$1x+2x-7 < 6\\\\3x-7 < 6\\\\3x < 6+7\\\\3x < 13\\\\x < \frac{13}{3}\\\\x < 4 \frac{1}{3}$

The graph will have an open circle at 13/3 = 4&1/3 = 4.333 approx. The shading is to the left. No inequality sign flip happens because we divided both sides by a positive number.

Your choice of three solutions is correct. You can pick anything smaller than 4.3333

================================================

Problem 3

$9-2x \le 6x-15\\\\-2x-6x \le -15-9\\\\-8x \le -24\\\\x \ge \frac{-24}{-8} \ \text{ inequality sign flip}\\\\x \ge 3$