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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%

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I need help working this algebra 2 question 3/4 - 5/6R = 7/10

Brilliant_brown [7]

Answer:

R = 3/50

Step-by-step explanation:

If you are trying to solve for a variable,

1. Simplify both sides of the equation:

3/4 - 5/6R = 7/10 will now be (-5/6r) + 3/4 = 7/10

2. Add -3/4 from both sides of the equation:

(-3/4) + 3/4 = 0 and (-3/4) - 7/10 = -1/20.

You will be left with: -5/6r = -1/20.

3. Divide both sides by (-5/6).

(-5/6r) / -5/6 = 1R or just R. (-1/20) / -5/6 = 3/50

You are left with the answer:

R = 3/50

I hope this helps!

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3 years ago

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Tpy6a [65]

Answer: 3, 4

Step-by-step explanation:

and the common factor between these 2 numbers is 1 so 3,4 is its greatest factor

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3 years ago

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introvert

Setler [38]

Answer:

$P(x \ge 5) = 1.000$ ---- At least 5 from marketing departments are extroverts

$P(x=15) = 0.013$ ---- All from marketing departments are extroverts

$P(x = 0) = 0.002$ ---------- None from computer programmers are introverts

Step-by-step explanation:

See comment for complete question

The question is an illustration of binomial probability where

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$(a):\ P(x \ge 5)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

Using the complement rule, we have:

$P(x \ge 5) = 1 - P(x < 5)$

So, we have:

$P(x < 5) =P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)$

$P(x = 0) = ^{15}C_{0} * (75\%)^0 * (1 - 75\%)^{15 - 0} = 1 * 1 * (0.25)^{15} = 9.31 * 10^{-10}$

$P(x = 1) = ^{15}C_{1} * (75\%)^1 * (1 - 75\%)^{15 - 1} = 15* (0.75)^1 * (0.25)^{14} = 4.19 * 10^{-8}$

$P(x = 2) = ^{15}C_{2} * (75\%)^2 * (1 - 75\%)^{15 - 2} = 105* (0.75)^2 * (0.25)^{13} = 8.80 * 10^{-7}$

$P(x = 3) = ^{15}C_{3} * (75\%)^3 * (1 - 75\%)^{15 - 3} = 455* (0.75)^2 * (0.25)^{12} = 0.0000153$

$P(x = 4) = ^{15}C_{4} * (75\%)^4 * (1 - 75\%)^{15 - 4} = 1365 * (0.75)^4 * (0.25)^{11} = 0.000103$

So, we have:

$P(x < 5) = (9.31 * 10^{-10}) + (4.19 * 10^{-8}) + (8.80 * 10^{-7}) + 0.0000153 + 0.000103$

$P(x < 5) = 0.00011922283$

Recall that:

$P(x \ge 5) = 1 - P(x < 5)$

$P(x \ge 5) = 1 - 0.00011922283$

$P(x \ge 5) = 0.9998$

$P(x \ge 5) = 1.000$ --- approximated

$(b)\ P(x = 15)$

$n = 15$ --- marketing personnel

$p = 75\%$ --- proportion that are extroverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x=15) = ^{15}C_{15} * 0.75^{15} * (1 - 0.75)^{15-15}$

$P(x=15) = 1 * 0.75^{15} * (0.25)^{0$

$P(x=15) = 0.013$

$(c)\ P(x = 0)$

$n=5$ ---------- computer programmers

$p = 70\%$ --- proportion that are introverts

So, we have:

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x = 0) = ^{5}C_0 * (70\%)^0 * (1 - 70\%)^{5-0}$

$P(x = 0) = 1 * 1 * (0.30)^5$

$P(x = 0) = 0.002$

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2 years ago

.Jim opened an account with $400. The account pays three percent quarterly. How much is in the account at the end of two years?

podryga [215]

We may have this planned like this:
two years is 8 quarters
Assuming simple interest, 
You will use the formula
A = P(1+rt) 
= 400(1+8*.03) 
= 496
I think this is the answer:) Hope it helps a lot

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3 years ago

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Doss [256]

Answer:

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Step-by-step explanation:

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