What is the probablity for the number of times Jada will pull a tile wit a star?
Answer: Number of tiles pulled out of the bag by Jada = 10, so she completed 10 trials.
The outcome of each trial was star, circle or a square.
Step-by-step explanation: We can find the experimental probability of pulling each shape from the bag based on the outcomes of Jada's experiment and number of trial.
- P(Star) = Number of stars/Number of trials= 5/10
- P(Circles)= Number of circles/Number of trials =3/10
- P(Square) = Number of Square/Number of trials = 2/10
Here, P refers to Probability of an outcome
If Jada repeats the experiment, She would likely again pull more stars than circles or square. She may get different number of stars,circles or squares but over a large number of trials, she would expect the ratio of the stars in the proportion. We can use this proportonal relationship to get the reasonable prediction about te number of times she pull different shape tiles.
<u>Number of outcomes(10 trials) </u> = <u>Number of outcomes(100 trials)</u>
Number of trials(10 trials) Number of trials (100 trials)
- Numbeof stars = 5/10 =s/100
5*10/ 10*10 =50/100 - Number of Circle = 3/10 = c/100
3*10/10*10 = 30/100 - Number of square = 2/10 = q/100
2*10/10*10 = 20/100
If she does the experiment 100 times then she should expect to pull out 50 stars tile, 30 circles and 20 Square.
Hence, Jada pulled out the 100 tiles from the bag.
Learn more about geometry questions here-: brainly.com/question/17140560
So firstly, add px onto both sides of the equation: 
Next, add 2 to both sides of the equation: 
Next, factor out -x on the right side of the equation: 
Next, divide both sides by (8 - p): 
Lastly, multiply both sides by -1 and <u>your answer will be 
</u>
Answer:
(A)12-19
Step-by-step explanation:
To construct a frequency distribution, we follow the steps below.
1. Find the largest and smallest values.
91 and 12 respectively
Compute the Range = Maximum - Minimum=91-12=79
2. Determine the number of classes desired.
In this case, we desire 10 classes
3. Next, we find the class width by dividing the range by the number of classes and rounding up.
79÷10 = 7.9
Rounded Up, Class Width=8
4. Pick a suitable starting point less than or equal to the minimum value.
Since our minimum value is 12, we pick a class starting with 12 and give a class width of 8.
Therefore the upper and lower limits of the first class is given as:
12-19.
Answer:
D. x=-3/2.
Step-by-step explanation:
4x÷2x=-3
2x=-3
x=-3/2
