All categories

3 years ago

Interpret the following table about school dropout at the basic level (5 marks)

Mathematics

1 answer:

barxatty [35]3 years ago

8 0

Answer:

Step-by-step explanation:

Given the following :

- - - Jhs

Basic level - - - - Boys - - - - Girls - - - - Total

Primary - - - - - - - 49 - - - - - - 51 - - - - - - 100%

Jhs - - - - - - - - - - 56 - - - - - - 44. - - - - - - 100%

With the information above,

Primary Dropout percentage:

BOYS : [49 / (49 +51)] × 100.= 49%

GIRLS : [51 / (49 +51)] × 100.= 51%

Jhs: Dropout percentage

BOYS : [56 / ( 56 + 44) × 100 = 56%

GIRLS : [44 / (44 +56)] × 100.= 44%

You might be interested in

In how many ways can a

Tomtit [17]

Answer:

The answer is 364. There are 364 ways of choosing a recorder, a facilitator and a questioner froma club containing 14 members.

This is a Combination problem.

Combination is a branch of mathematics that deals with the problem relating to the number of iterations which allows one to select a sample of elements which we can term "r" from a collection or a group of distinct objects which we can name "n". The rules here are that replacements are not allowed and sample elements may be chosen in any order.

Step-by-step explanation:

Step I

The formula is given as

$C (n,r) = \frac{n}{r} = \frac{n!}{(r!(n-r)!)}$

n (objects) = 14

r (sample) = 3

Step 2 - Insert Figures

C (14, 3) = $(\frac{14}{3})$ = $\frac{14!}{(3!(14-3)!)}$

= $\frac{87178291200}{(6 X 39916800)}$

= $\frac{87178291200}{239500800}$

= 364

Step 3

The total number of ways a recorder, a facilitator and a questioner can be chosen in a club containing 14 members therefore is 364.

Cheers!

6 0

3 years ago

What is 2+2 in math and the answer is 4

Marianna [84]

Answer:

Step-by-step explanation:

7 0

2 years ago

Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate

kolbaska11 [484]

Answer:

You can proceed as follows:

Step-by-step explanation:

First solve the quadratic inequality $x^{2}-5x+4\leq 0$ . To do that, factorize, then we have that $(x-4)(x-1)\leq 0$ . This implies that

$x-1\leq 0\, \text{and}\, x-4\geq 0$

$x-1 \geq 0\, \text{and}\, x-4\leq 0$

In the first case the solution is the empty set $\emptyset$ . In the second case the solution is the interval $1\leq x \leq 4$ . Now we have that

$A=[1,4]$

$B=(3,5)$

$C=(3,4]$ .

To show that $A\cap B\subseteq C$ consider $x\in A\cap B$ . Then $1\leq x \leq 4\, \text{and}\, 1$ , this implies that $3$ , then $x\in C$ . Now, to show that $C\subseteq A\cap B$ consider $x\in C$ , then $3$ , then $1\leq x \leq 4\, \text{and}\, 3$ , then $x\in [1,4] \, \text{and}\, x\in (3,5)$ , this implies that $x\in A\cap B$ .