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3 years ago

In a classroom of 33 students, the ratio of boys to girls is 3:8. How many boys are in the class?

Mathematics

2 answers:

chubhunter [2.5K]3 years ago

7 0

the answer is 9

i'm only writing more because its making me or else i can't post this. I need the points badly! so... yeah... also thanks to the person who posted before me because i thought you were wrong until i read the step by step

hope you pass your Imagine Math lesson or whatever program you are doing!

good luck!

--- a weaboo who really wants to be an anime character, and is also in love with the show Yuri!!! on ICE

Hunter-Best [27]3 years ago

3 0

X=number of boys
y=number of girls.

We can suggest the following system of equations:
x+y=33
x/y=3/8

We solve this system of equations by substitution method.
x/y=3/8 ⇒x=3y/8

(3y/8)+y=33
least common multiple=8
3y+8y=33*8
11y=33*8
y=(33*8)/11
y=3*8
y=24

x+y=33
x+24=33
x=33-24
x=9

number of boys /total number of students=9/33=3/11

There are 9 boys and 24 girls in the class.

Answer: C: 9. since 3/11 of the students are boys.

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3 years ago

Help please. <br> Give the slope of y - 3 = 4(x + 8) and a point on the line

Sauron [17]

-----------------------------------------------------------
Make y the subject :
-----------------------------------------------------------
y - 3 = 4(x + 8)
y - 3 = 4x + 32 // Apply distributive property
y = 4x + 35 // Add 3 to both sides

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Identify Slope :
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Slope = 4

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Find a point on the line :
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When x = 0,
y = 4(0) + 35 // Sub x = 0 into the equation
y = 35 // Combine like terms
Point = (0, 35)

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Answers :
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3 years ago

Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe

nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let $D^c$ be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that one person in 10,000 people has a rare genetic disease.

So, $P(D)=\frac{1}{10000}$

Only 0.4% of the people who don't have it test positive.

$P(A|D^c)$ = probability that a person is tested positive given he don't have a disease = 0.004

$P(D^c)=1-\frac{1}{10000}$

Formula: $P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}$

$P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}$

P(D|A)= $\frac{2470}{2471}$ =0.9995

P(D|A)= $0.9995$

A)The probability that someone who tests positive has the disease is 0.9995

(B)

$P(D^c|A^c)$ =probability that someone does not have disease given that he tests negative

$P(A^c|D^c)$ =probability that a person tests negative given that he does not have disease =1-0.004

=0.996

$P(A^c|D)$ =probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: $P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}$

$P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}$

$P(D^c|A^c)=0.99999$

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0

3 years ago