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blagie [28]
3 years ago
11

In a certain​ fraction, the denominator is less 7 than the numerator. If 3 is added to both the numerator and​ denominator, the

resulting fraction is equal to 15/8
. Find the original fraction.
Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
7 0

Answer:

original fraction =12/5

Step-by-step explanation:

let x = numerator

therefore

   x+3/x-7+3 =15/8

    x+3/x-4 =15/8

      8(x+3)=15(x-4)

      8x+24=15x-60

       84=7x

        84/7=x

        12=x

therefore original fraction = 12/5

You might be interested in
Someone please help! find x round to the nearest tenth! trigonometry!​
TEA [102]

Answer:

<h2>x = 745.5 ft</h2>

Step-by-step explanation:

Look at the picture.

Use sine:

sine=\dfrac{opposite}{hypotenuse}

We have a measure of the angle and length of the opposite:

\alpha=28^o,\ opposite=350\ ft

\sin28^o\approx0.4695

Substitute:

\dfrac{350}{x}=0.4695

\dfrac{350}{x}=\dfrac{4695}{10000}            <em>cross multiply</em>

4695x=3500000           <em>divide both sides by 4695</em>

x=\dfrac{3500000}{4695}\to x\approx745.5

5 0
3 years ago
A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created that is made up of 3 senior o
Zigmanuir [339]

Answer:

80

Step-by-step explanation:

A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created that is made up of 3 sr officers and 1 jr officer, how many different committees are possible?

8

24

58

80*

210

I initially got this one wrong, but after I saw the answer which is 80, I was able to solve it. But im not sure if this is the proper way of solving it, or if someone can suggest a more efficient way of solving it?

This is what I did, which was wrong:

Sr Officers + Jr Officers

(6!/3! 3! = 20) + (4!/3! = 4) = 24

It seems like the correct answer is:

Sr Officers X Jr Officers

(6!/3! 3! = 20) x (4!/3! = 4) = 80

If this is the most efficient way of solving, when do you know when to add them as opposed to multiplying them in this case?

7 0
2 years ago
Suppose that the waiting time for a license plate renewal at a local office of a state motorvehicle department has been found to
zysi [14]

The probability that a randomly selected individual will have a waiting time between 16 and 44 minutes is 93.72%.

Given mean of 30 minutes and standard deviation of 7.5 minutes.

In a set with mean d and standard deviation d. , the z score is given as:

Z=(X-d)/s.

where d is sample mean and s is standard deviation.

We have to calculate z score and then p value from normal distribution table.

We have been given d=30, s=7.5

p value of Z when X=44 subtracted by the p value of Z when X=16.

When X=44,

Z=(44-30)/7.5

=14/7.5

=1.87

P value=0.9686

When X=16

Z=(16-30)/7.5

=-1.87

P Value=0.0314.

Required probability is =0.9686-0.0314

=0.9372

=93.72%

Hence the probability that a randomly selected individual will have a waiting time between 16 and 44 minutes is 93.72%.

Learn more about z test at brainly.com/question/14453510

#SPJ4

7 0
2 years ago
Muna claims that the expression (8)(16) + (8)(12) + (16)(12) represents the
aivan3 [116]

The surface area of a right rectangular prism is the amount of space on it.

Muna's claim is incorrect

<h3>How to determine the surface area?</h3>

The surface area of a right rectangular prism is calculated as:

A = 2 * (lw + lh + wh)

So, we have:

A = 2 * (8 * 16 + 8 * 12 + 16 * 12)

The above expression implies that Muna's claim is incorrect

Read more about surface areas at:

brainly.com/question/27002333

6 0
2 years ago
A basin is filled by two pipes in 12 minutes and 16 minutes respectively. Due to the obstruction of water flow after the two pip
olasank [31]

Answer:

The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes

Step-by-step explanation:

The given information are;

The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes

The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes

Therefore, the flow rate of pipe A = V/12

The flow rate of pipe B = V/16

Due to the restriction, we have;

The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity

The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity

Whereby the tank is filled 3 minutes after the restriction is removed, we have;

\dfrac{7}{8} \times \dfrac{V}{12} \times t + \dfrac{5}{6} \times \dfrac{V}{16} \times t +  \dfrac{V}{12} \times 3 + \dfrac{V}{16} \times 3 = V

Simplifying gives;

\dfrac{(2\cdot t +7) \cdot V}{16}  = V

2·t + 7 = 16

t = (16 - 7)/2 = 4.5 minutes

Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal

7 0
3 years ago
Read 2 more answers
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