Answer:

Step-by-step explanation:
Given




Required
The rate of change (b)
The above graph is represented as:

For:
;
We have:





For
,
We have:



Substitute 

Divide by 2


<em>Hence, the rate of change is 2.50</em>
Answer:
1/6
Step-by-step explanation:
there are 6 dresses altogether and 1 blue dress.
Umm..... according to what I know <span> </span><span>The ratio of the sides of a 30°-60°-90° triangle are:
short leg : long leg : hypotenuse
1 : √3 : 2
</span><span> if the short leg is 7, the long leg is √3 times that:
Answer:
7√3 ≈ 12.12 inches </span>
Answer:
52.9 inches .
Step-by-step explanation:
Given that the triangular indentation has an area of 100 in.² and the base and height of this traingle are represented by expressions 3x and x+3 respectively . We need to find out the <u>perimeter </u>to the nearest tenth.
As we know that the area of triangle is ,
Substituting the respective values,
On using the <u>Quadratic</u><u> formula</u> , we have;

On simplifying above , we will get ,

Since sides can't be negative, therefore,

Therefore ,
Next let's find out the <u>hypotenuse </u>using Pythagoras theorem, as ;
Now we may find perimeter as ,

<span>1. Suppose that a family has an equally likely chance of having a cat or a dog. If they have two pets, they could have 1 dog and 1 cat, they could have 2 dogs, or they could have 2 cats.
What is the theoretical probability that the family has two dogs or two cats?
25% chance
</span><span>2. Describe how to use two coins to simulate which two pets the family has.
</span>
You could use the coins to simulate which pet the family has by flipping them and having head be dog and tails be cat (or vice-versa).
<span>3. Flip both coins 50 times and record your data in a table like the one below.
</span><span>Based on your data, what is the experimental probability that the family has two dogs or two cats?
</span>
Based on the results, I concluded that for Heads, Heads (which could be dogs or cats) there was a 24% chance and for Tails, Tails there was a 26% chance
<span>4. If the family has three pets, what is the theoretical probability that they have three dogs or three cats?
1/8 chance (accidentally messed up there) or 12.5%
</span><span>5. How could you change the simulation to generate data for three pets?
</span><span>
To flip 3 coins and add more spots on the chart.
I hope that this helps because it took a while to write out. If it does, please rate as Brainliest
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