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maria [59]
3 years ago
6

Please help ASAP I’m not good at math

Mathematics
2 answers:
kramer3 years ago
7 0

Answer: B

12/34 sold are recliners. This can be reduced to 6/17, which is your final answer.

SpyIntel [72]3 years ago
4 0
Answer: the answer is A
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25 - so<br> Is the following number rational or irrational?
svlad2 [7]

Answer:

25 is a rational number

Step-by-step explanation:

This is because, a rational number needs to be in p/q form, where q ≠ 0

This is possible for 25, 25 in p/q form =

25/1

If my answer helped, kindly makr me as the brainliest!!

Thank You!!

4 0
3 years ago
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Which value is equivalent to 7 multiplied by 3 multiplied by 2 whole over 7 multiplied by 5, the whole raised to the power of 2
Valentin [98]

Answer:

i think it might be A

Step-by-step explanation:

6 0
3 years ago
0.31 repeating as a fraction
sukhopar [10]
I believe its 31/100! <span>To write 0.3111 as a fraction you have to write 0.3111 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.</span>

7 0
3 years ago
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Geometry math question no Guessing and Please show work thank you
vesna_86 [32]

First let us find the length of JN

JN =JK +KN

JN = 82+105 = 187

JN=187..............(1)

UC= JN -( JH +HU+CN)

We are given :

JH= 22, HU = 96 and CN = 51

plugging all the values we get

UC = 187-( 22+96+51)

UC =187 -169

UC = 18

Answer is UC =18 ( option B)

8 0
4 years ago
A bin contains 25 light bulbs, 5 of which are in good condition and will function for at least 30 days, 10 of which are partiall
Ira Lisetskai [31]

Answer:

The probability that it will still be working after one week is \frac{1}{5}

Step-by-step explanation:

Given :

Total number of bulbs = 25

Number of bulbs which are good condition and will function for at least 30 days = 5

Number of bulbs which are partially defective and will fail in their second day of use = 10

Number of bulbs which are totally defective and will not light up = 10

To find : What is the probability that it will still be working after one week?

Solution :

First condition is a randomly chosen bulb initially lights,

i.e. Either it is in good condition and partially defective.

Second condition is it will still be working after one week,

i.e. Bulbs which are good condition and will function for at least 30 days

So, favorable outcome is 5

The probability that it will still be working after one week is given by,

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

\text{Probability}=\frac{5}{25}

\text{Probability}=\frac{1}{5}

5 0
3 years ago
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