Answer:
Length of the field: 94 m
Width of the painting: 61 cm
Step-by-step explanation:
Use the perimeter formula, P = 2l + 2w, to find the length:
Plug in the perimeter and width into the equation:
P = 2l + 2w
336 = 2l + 2(74)
336 = 2l + 148
188 = 2l
94 = l
So, the length of the field is 94 m.
To find the width of the painting, use the area formula, A = lw
Plug in the area and length into the equation:
A = lw
5795 = 95w
61 = w
So, the width of the painting is 61 cm.
Length of the field: 94 m
Width of the painting: 61 cm
Answer:
x is less than or equal to negative 14.
Step-by-step explanation:
so for now we should pretend that the greater than or equal to is an equal sign, and simplify the problem to get x alone on a side
remove 5 from each side
-x/3=1/3-5
-x/3=14/3
multiply both sides by 3
-x=14
x=-14
so we can now replace the equal sign with the equal or greater to, so x is less than or equal to -14
Hey there! I am on the same one. :) I will help you out a little.
<span>Assume that all six outcomes of a six-sided number cube have the same probability. What is the theoretical probability of each roll?
• 1: 1/6
• 2: 2/6
• 3: 3/6
• 4: 4/6
• 5: 5/6
• 6: 6/6
</span>
<span>Using the uniform probability model you developed, what is the probability of rolling an even number?
1/6 Roll a number cube 25 times. Record your results here.
</span><span>
<span><span>
<span>
<span>1st
toss=</span>6</span>
</span>
<span>
<span>
<span>2nd
toss=</span>4</span>
</span>
<span>
<span>
<span>3rd
toss=</span>6</span>
</span>
<span>
<span>
<span>4th
toss=</span>6</span>
</span>
<span>
<span>
<span>5th
toss=</span>3</span>
</span>
<span>
<span>
<span>6th
toss=</span>3</span>
</span>
<span>
<span>
<span>7th
toss=</span>4</span>
</span>
<span>
<span>
<span>8th
toss=</span>2</span>
</span>
<span>
<span>
<span>9th
toss=</span>6</span>
</span>
<span>
<span>
<span>10th
toss=</span>5</span>
</span>
<span>
<span>
<span>11th
toss=</span>1</span>
</span>
<span>
<span>
<span>12th
toss=</span>4</span>
</span>
<span>
<span>
<span>13th
toss = </span>5</span>
</span>
<span>
<span>
<span>14th
toss =</span>1</span>
</span>
<span>
<span>
<span>15th
toss=</span>4</span>
</span>
<span>
<span>
<span>16th
toss=</span>2</span>
</span>
<span>
<span>
<span>17th
toss=</span>2</span>
</span>
<span>
<span>
<span>18th
toss=</span>2</span>
</span>
<span>
<span>
<span>19th
toss=</span>6</span>
</span>
<span>
<span>
<span>20th
toss=</span>5</span>
</span>
<span>
<span>
<span>21st
toss=</span>3</span>
</span>
<span>
<span>
<span>22nd
toss=</span>4</span>
</span>
<span>
<span>
<span>23rd
toss=</span>3</span>
</span>
<span>
<span>
<span>24th
toss=</span>3</span>
</span>
<span>
<span>
25
toss=5
How
many results of 1 did you have? __2____________ How
many results of 2 did you have? ____4__________ How
many results of 3 did you have? ____5__________ How
many results of 4 did you have? ______5________ How
many results of 5 did you have? ______4________
How
many results of 6 did you have? ______5________
Based
on your data, what is the experimental probability of each roll?
<span>
1. 2/25 or 0.08
2. 4/25 or 0.16
3. 5/25 or 0.24
4. 5/25 or 0.2
5.4/25 or 0.16
<span>
6. 5/25 or 0.2</span></span>Using
the probability model based on observed frequencies, what is the probability of
rolling an even number?
3/6 = ½ or 0.5
Was your experimental probability
different than your theoretical probability? Why or why not?
<span>It somewhat is! The
denominator is 25 for the experimental probability, and 6 for the theoretical
probability.</span><span>
</span><span>Have a lovely day! Cheerio. :) </span></span>
</span>
</span></span>
With these, always write out the multiples first.
Start like this:
(assume one of the factors is negative)
1 and 1120
2 and 560
4 and 280
5 and 224
7 and 160
8 and 140
10 and 112
14 and 80
16 and 70
20 and 56
28 and 40
32 and 35
from those, the obvious choice is the one with a difference of three. In this case, 32 and 35, because -32 + 35 equals 3.
