1 kilogram of baking soda costs $4.5
Step-by-step explanation:
To solve this problem, we will follow the steps bellow:
Using proportion;
Let x be the cost of 1 kg of baking soda
kilogram = $ 2
1 kilogram = x
Cross - multiply
× x = $2 × 1
× x = $2
= $2
Multiply both-side of the equation by 9
× 9= $2 × 9
At the left-hand side of the equation 9 will cancel-out 9, hence our equation becomes
4x = $18
Divide both-side of the equation by 4
= $
x = $4.5
1 kilogram of baking soda costs $4.5
Step-by-step explanation:
In case of dogs the value 10 is the minimum value. So all the values lie above 10. In total there were 100 dogs.
So for dogs, we can say number of dogs above the value of 10 pound are 100.
In case of Cats, 10 lies at the position of median. Median is the central value and 50% values lie above the median value. So number of cats with weight above 10 pound is 50.
Thus, we can conclude that there were 50 more dogs than the cats with weight over 10 pounds. So option C gives the correct answer.
Answer:
a. The equation is 33+m
b. 333 dollars
Step-by-step explanation:
Deposited money:
March: 120
April: 120+(120×20)/100=120+24=144
May: 144+(144×20)/100=144+18,8=162,8
June: 162,8+(162.8×20)/100=162,8+32.56=195.36
July: 195.36+(195.36×20)/100=195.36+23.87=219.23
August: 219.23+(219.23×20)/100=219.23+43.85=263.08
120+144+162.8+195.36+219.23+263.08=1105.19
So the answer is no.
Answer: i believe that it is 49 hits
Comment:
and believe me sis i feel ya i'm in 8th grade and fell behind tryna help other people because i don't want others to feel what i felt.
GOOD LUCK!
