Let's say the first truck weighs x tons
Then, the weight of 2nd truck = x+2 tons
The weight of 3rd truck = (x + 2) + 2 = x+4 tons
The weight of 4th truck = (x + 4) + 2 = x+6 tons
Total weight of 4 trucks:
x + (x+2) + (x+4) + (x+6) = 32
which can be solved easily to give x = 5
Answer:
Step-by-step explanation:
in tri ADC and tri BDC
∠ADC =∠BDC = 90
DC is common
AD = BD (given)
triangle ADC ≅ tri BDC by SAS congruency
hence AC = BC by CPCT ( congruent parts of congruent triangles)
hence, BC = 13
Answer:
Exponential Growth, 1200(1.15)
Step-by-step explanation:
It is a Growth function because it is "increasing" yearly. The equation is multiplying with 1.15 because when it is increasing, you need to add 1 to the percentage rate.
20 / 27 is the probability that a student chosen randomly from the class passed the test or completed the homework.
<u>Step-by-step explanation:</u>
To find the probability that a student chosen randomly from the class passed the test or complete the homework :
Let us take,
- Event A ⇒ a student chosen randomly from the class passed the test
- Event B ⇒ a student chosen randomly from the class complete the homework
We need to find out P (A or B) which is given by the formula,
⇒ P (A or B) = P(A) + P(B) - P(A∪B)
<u>From the given table of data,</u>
- The total number of students in the class = 27 students.
- The no.of students passed the test ⇒ 15+3 = 18 students.
P(A) = No.of students passed / Total students in the class
P(A) ⇒ 18 / 27
- The no.of students completed the homework ⇒ 15+2 = 17 students.
P(B) = No.of students completed the homework / Total students in the class
P(B) ⇒ 17 / 27
- The no.of students who passes the test and completed the homework = 15 students.
P(A∪B) = No.of students both passes and completes the homework / Total
P(A∪B) ⇒ 15 / 27
Therefore, to find out the P (A or B) :
⇒ P(A) + P(B) - P(A∪B)
⇒ (18 / 27) + (17 / 27) - (15 / 27)
⇒ 20 / 27
∴ The P (A or B) is 20/27.
