Evaluate the expression when y = 5. 3 + y + 6 9 14 19 21

Tion 15/30

vitfil [10]

The percentage of students that passed the exam is 25.2%

The given parameters;

number of the males, = 85

number of the females, = 30

percentage of the female that passed = 40%

percentage of the male that passed, = 20%

To find:

the percentage of students that passed the exam

The number of the male students that pass the exam;

$= 0.2 \times 85 \\\\= 17$

The number of the female students that pass the exam;

$= 0.4 \times 30 \\\\= 12$

The total number of students in the school;

= 85 + 30

= 115

The total number of the students that passed the exam;

= 17 + 12

= 29

The percentage of students that passed the exam;

$= \frac{total \ number \ that \ passed }{total \ number \ of \ the \ students} \times 100\% \\\\= \frac{29}{115} \times 100\%\\\\= 25.2 \%$

Thus, the percentage of students that passed the exam is 25.2%

Learn more here: brainly.com/question/1163846

6 0

3 years ago

4. At a local university 54.3% of incoming first-year students have computers. If 3 students are selected at random, and the fol

valentinak56 [21]

Answer:

a) 0.0954

b) 0.9045

c) 0.1601

Step-by-step explanation:

We are given the following information:

We treat first-year students having computers as a success.

P(first-year students have computers) = 54.3% = 0.543

Then the number of first year students follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 3

a) P(None have computers)

$P(x =0) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3\\\\= 0.0954$

b) P(At least one has a computer)

$P(x \geq 1) \\\\= \binom{3}{0}(0.543)^0(1-0.543)^3+\binom{3}{1}(0.543)^1(1-0.543)^2+\binom{3}{3}(0.543)^3(1-0.543)^0\\\\=0.3402+0.4042+0.1601= 0.9045$

c) P(All have computers)

$P(x =3) \\\\= \binom{3}{3}(0.543)^3(1-0.543)^0\\\\= 0.1601$

3 0

3 years ago

In a triangle ABC, AD is drawn perpendicular to BC. Prove that AB2 - BD2 = AC2 - CD2.

Finger [1]

Answer:

Given: In triangle ABC , AD is drawn perpendicular to BC.

Since AD is drawn perpendicular to BC, it creates two right triangles: ADB and ADC.

Prove that: $AB^2-BD^2 = AC^2-CD^2$

Pythagoras triangle for right angle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.

In a right angle triangle ADB;

$AD^2+BD^2 = AB^2$ [By Pythagoras theorem]

or

$AD^2= AB^2-BD^2$ .......[1]

Now, in right angle triangle ADC;

$AD^2+CD^2 = AC^2$ [By Pythagoras theorem]

or we can write this as;

$AD^2= AC^2-CD^2$ ......[2]

Substituting the equation [1] in [2] we get;

$AB^2-BD^2 =AC^2-CD^2$ hence proved!

5 0

3 years ago

Date:

Free_Kalibri [48]

Answer:

Yes

Step-by-step explanation:

We know that 1.39 is less than 1.5 and 3.75 is less than 4.

Hence the price of 3.75kg is going to be less than: 4*1.5=$6

Given that apples cost $1.39 per kilogram, then the price of 3.75 kg of apples is 3.75*1.39=$ 5.2125.

Because 5.2125 is actually less than $6

This means that $6 can buy 3.75 kg of apples.

3 0

3 years ago

What is the complementary angle of an angle that measures 30 degrees?

alina1380 [7]

60 degree is the complementary angle of 30 degree

4 0

3 years ago