Answer:
x = 112°, y = 68°
Step-by-step explanation:
68° and y are corresponding angles and are congruent, then
y = 68°
x and y are adjacent angles and are corresponding ( sum to 180° ), then
x + y = 180°
x + 68° = 180° ( subtract 68° from both sides )
x = 112°
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.
ok. We are given the following equation:
Where:
Since we are asked to determine "H" we will solve for "H" in the equation. To do that we will first multiply both sides by -1:
Now, we will use the following property of logarithms:
Applying the property and having into account that:
We get:
Now we substitute the given value of "pH = 4.5":
Solving the operation:
Therefore, the hydrogen ion concentration of the fluid is 0.000032
First do -x then -x to 4x so you have 3x-6=9.
Add 6 to -6 to cancel it out, then add 6 to 9.
3x-6. Divide 3from x, divide 3from 6. X=2. Plz answer my question too plz
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