Answer:

Step-by-step explanation:
Alvin takes pictures with his smartphone.
Of the pictures
are animals.
of the animals are dogs.
To find which fraction of all pictures are pictures with dogs, you can simply multiply two fractions:

<u>Another way</u> to solve this question:
Let the number of all pictures be 24 (divisible by both 4 and 6).
Then the number of pictures with animals is

and the number of pictures with dogs is

Hence, the fraction of pictures with dogs of all pictures is

The range is [4,∞)
Or from 4 to positive infinity
Or y is greater than or equal to 4
Answer:
The cooking club sales covers the expenditure when 2 piece of cakes are sold.
Step-by-step explanation:
Given:
Selling price of each piece of cake = $10
Cost for booth at fair = $10
Ingredients for each piece of cake = $5
We need to find the number of pieces of cake sold when the sales cover the expenditures.
Solution:
Let the number of pieces be 'x'
So We can say that the point at which the sales cover the expenditures can be calculated as Selling price of each piece of cake multiplied by number of pieces will be equal to Cost for booth at fair plus Ingredients for each piece of cake multiplied number of piece of cakes.
framing in equation form we get;

Now Subtracting both side by '5x' using Subtraction property we get;

Now Dividing both side by 5 we get;

Hence The cooking club sales covers the expenditure when 2 piece of cakes are sold.
to find the hypotenuse (or the ramp in question 4) you can use the equation

with a and b being the 3 feet and 9 feet and c being the unknown (ramp).
So if you plug 3 and 9 into the equation, it looks like this.

then square them.

simplify.

take the square root of both sides.

is approx. 9.5 so the answer would be H.
Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
_____
My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)