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3 years ago

Divide and write in simplest form 2 3/8 divided by 1 3/4

Mathematics

1 answer:

s344n2d4d5 [400]3 years ago

7 0

First, change each mixed fraction into improper

2 3/8 = 19/8

1 3/4 = 7/4

The problem will look like: (19/8)/(7/4)

To solve, flip the second fraction, and change the division sign into a multiplication sign.

(19/8)/(7/4) = (19/8) x (4/7)

Multiply across

(19/8) x (4/7) = 76/56

Simplify.

76/56 simplified = 19/14

Change to mixed fraction

19/14 = 14/14 + 5/14 = 1 5/14

1 5/14 is your answer

hope this helps

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Which values of m and b will create a system of equations with no solution? Select two options.

Roman55 [17]

Answer:

2nd and 5th options

Step-by-step explanation:

the solution to a system of equations is at the point of intersection of the 2 lines.

If the lines are parallel then they never intersect and will have no solution.

The slopes m of parallel lines are equal.

The options with equal slopes , both - 2 are the

2nd and 5th options

6 0

2 years ago

Please Help I don’t get it at all N/2+5=3

eduard

N/2+5=3

move +5 to the other side

sign changes from +5 to -5

N/2+5-5= 3-5

N/2= -2

Mutiply both sides by 2/1

N/2*2/1 - Cross out 2 and 2 , divide by 2 then becomes N

-2(2)/1= -4/1= -4

N= -4

Answer : N= -4

5 0

3 years ago

Read 2 more answers

The population of a town was 6,000

Wewaii [24]

Answer: It is expected to increase by 240 people

Step-by-step explanation: 0.04*6000=240

5 0

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

CAN SOMEONE PLEASE HELP!!!

Andreas93 [3]

Step-by-step explanation:

AC=AB+BC-cos. AC=100+66-cos. AC=166-0.906. AC=√165.094. AC=12.8

4 0

2 years ago