Answer:
126
Step-by-step explanation:
To solve this, we need to use PEMDAS: Parentheses, exponents, multiplication, division, addition, subtraction
((4 x 8) + 10) x 3
Parentheses - multiplication
(32 + 10) x 3
Parentheses - addition
42 x 3
multiplication
126
To find out the result, we have to divide 365 days by 28 days:
365 / 28 = 13,03571428571429
So we can see it's 13 lunar months with something extra. To find that something extra we can multiply 28 by 13 and then substract the result from 365:
365 - (13 * 28) =
= 365 - 364 =
= 1
So there are 13 lunar months and one day.
22 students in each class. (88 divided by 4=22)
Answer:
There is a 100% probability that the selected tick is also a carrier of Lyme disease.
Step-by-step explanation:
The problem states that:
Of the ticks that carry at least one of these diseases in fact carry both of them.
So, if a tick carries one of these diseases, there is a 100% probability that is carries another.
What is the probability that the selected tick is also a carrier of Lyme disease?
There is a 100% probability that the selected tick is also a carrier of Lyme disease.
Answer:
Step-by-step explanation:
<u>Given quadratic function:</u>
<u>Points on the graph:</u>
- (-2,-35), (1,-5), (3,- 15)
<u>Substitute values of x and y and solve the system of equations:</u>
- -35 = a(-2)² + b(-2) + c ⇒ -35 = 4a - 2b + c ⇔ eq 1
- -5 = a(1)² + b(1) + c ⇒ -5 = a + b + c ⇔ eq 2
- -15 = a(3)² + b(3) + c ⇒ -15 = 9a + 3b + c ⇔ eq 3
<u>Subtract eq 2 from eq 1:</u>
- -35 - (-5) = 4a - 2b + c - a - b - c
- -30 = 3a - 3b
- b = a + 10 ⇔ eq 4
<u>Subtract eq 2 from eq 3:</u>
- -15 - (-5) = 9a + 3b + c - a - b - c
- -10 = 8a + 2b
- b = -4a - 5 ⇔ eq 5
<u>Compare eq 4 and eq 5, solve for a:</u>
- a + 10 = -4a - 5
- a + 4a = -5 - 10
- 5a = -15
- a = -3
<u>Find the value of b using eq 4:</u>
<u>Find the value of c using eq 2:</u>
- -5 = -3 + 7 + c
- c = -5 - 4
- c = -9
<u>We now have a, b and c:, the function is:</u>
