Answer:
whats the question? its a little blurry and hard to see
Step-by-step explanation:
Answer:
A=6.72
R=2%
g(5)=215.04
Step-by-step explanation:
explanation is the picture.
HOPE THIS HELPED!!! HAVE A GREAT DAY!!!
A.) 7^4x = 10
log base 10 (7^4X) = log base 10 (10)
4x log base 10 (7) = 1
4x (0.8451) = 1
3.3804x = 1
x = 0.2958
b.) ln(2) + ln(4x-1) = 5
ln (2 * 4x-1) = 5
ln (8x-2) = 5
log base (3) (8x-2) = 5
e^5 = 8x-2
e^5+2 = 8x
x = 18.8016
To find the amount of solutions, you must find the discriminant. The discriminant is b^2 - 4ac. Number 6 has 2 solutions and number 7 has 1 solution.
<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
</span>
