1. Nasim thinks of a number.

solve this problem by using a 5d process or writing and solving an equation. No matter which method you use, be sure to define y

marysya [2.9K]

Answer:

width of the rectangle=x=4in and length of the rectangle=11in

Step-by-step explanation:

Let the width of the rectangle be x, then according to the question, length is one less than three times its width that is l=3x-1.

Now, width=b=x and length=l=3x-1, thus

Perimeter of the rectangle= 2(l+b)

⇒ $30=2(3x-1+x)$

⇒ $15=(3x-1+x)$

⇒ $15=4x-1$

⇒ $16=4x$

⇒ $x=4in$

Thus, width of the rectangle=x=4in and length of the rectangle=l=3x-1=3(4)-1=12-1=11in.

5 0

3 years ago

Solve each of the following equations. Be make sure to check your solutions. Solve for part b only.

Alona [7]

Answer:

or, 3x-5=2.5x+3-x+4

or, 3x-1.5x=7+5

or, 2.5x=12

or, x=4.8 solved.

3 0

3 years ago

4.7 in 8.8 in A) 20.68 in- C) 41.36 in- B) 10.3 in D) 21.48 in

alexandr402 [8]

Answer:

if you're multiplying them it's c

Step-by-step explanation:

4.7•8.8=41.36

6 0

2 years ago

Can you please help me out

Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

$\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}$

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,

$\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}$ $\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}$

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

$\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}$

Hence, the best option is G.

5 0

1 year ago

Misha used a probability simulator to pull 3 colored marbles from a bag and flip a coin 50 times. The results are shown in the t

kotykmax [81]

To solve this problem, let us first remember that probability is:

Probability = number of successful events / number of total events

Now, let us first calculate the probability of getting a blue marble. There are 20 blue marbles out of 50 total marbles, therefore the probability is:

P (blue marble) = 20 / 50 = 0.4

Next, let us calculate for the probability of landing tails up. There are 22 tails success out of 50 times the coin was tossed, therefore the probability is:

P (tails) = 22 / 50 = 0.44

Finally, since we want to find the Probability of getting blue and Probability of landing tails, we multiply the two:

P (blue and tails) = 0.4 * 0.44

P (blue and tails) = 0.176 (ANSWER)

Now if the choices are in fraction:

P (blue and tails) = (20 / 50) (22 / 50) = 440 / 2500

8 0

3 years ago

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