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2 years ago

Please i need help wha is 5/3 + 2/4

2 answers:

7 0

Using like denominators, we can turn this into 20/12 + 6/12, then you just add the numerators to get 26/12, then simplify to get 13/6

vampirchik [111]2 years ago

5 0

Answer:

26/12

simplify—>13/6

Step-by-step explanation:

Find the LCD (lowest common denominator)

The LCD of these two fractions is 12. Convert them to their denominators.

5/3 *4/4 = 20/12

2/4 *3/3= 6/12

Then add! 6/12 + 20/12 = 26/12

Simplify —> 13/6

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Algebra help? picture attached :)

galina1969 [7]

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3 0

2 years ago

Gareth 30 students in a class and the ratio of boys to girls in the class is 2:3 calculate the number of goals

pishuonlain [190]

Answer:

There are 12 boys and 18 girls

Step-by-step explanation:

Equations

There are 30 students in a class. Let's call:

x = number of boys

Since the sum of boys and girls is 30:

30 - x = number of girls

The ratio of boys to girls is 2:3, thus:

$\displaystyle \frac{x}{30-x}=\frac{2}{3}$

Crossing denominators:

3x = 2(30 - x)

Operating the parentheses:

3x = 60 - 2x

Adding 2x:

5x = 60

Dividing by 5:

x = 60/5 = 12

x = 12

There are 12 boys and 30-12=18 girls

8 0

2 years ago

7 9/16÷2 3/4 A. 214 B. 234 C. 414 D. 41316

sukhopar [10]

Answer:

Step-by-step explanation:

8 0

1 year ago

Evaluate the expression for x=5, y=3,and z=1/4. 5x−6y+20z/4yz

Anna11 [10]

Just substitute each number into the expression:
5(5) - 6(3) + 20(1/4)/4(3)(1/4) = 25 - 18 + 5/3 = 7 + 5/3 = 26/3 or 8 and 2/3.

4 0

3 years ago

Read 2 more answers

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

2 years ago