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DanielleElmas [232]
3 years ago
6

1/x + 1/3x =6 how do you solve this equation for x

Mathematics
1 answer:
Elodia [21]3 years ago
3 0

Answer:

x = 2/9 (or = 0.222)

Step-by-step explanation:

1/x + 1/(3x) =6  (multiply both sides by 3x)

(1/x) (3x)  + [1/(3x)](3x) =6 (3x)

3 + 1 = 18x

4 = 18x

x = 4/18

x = 2/9 (or = 0.222)

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Write this trinomial i’m factored form. 5s^2-13s+6
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(5s+2)(s-3) as writing it as 5s^2-15s+2s+6 is the same as the last form. But makes it easier to factorise as there are common factors.

3 0
3 years ago
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Rachel has a collection of 40 stuffed animals. 3/8 of the animals are bears and are dogs. How many bears and dogs does she have?
Ludmilka [50]

Answer:

total animal 40

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Step-by-step explanation:

hope it helped u buddy

5 0
3 years ago
A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

            \sigma = population standard deviation = 1.8 years

            n = sample of students = 36

            \mu = population average age

So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
3 years ago
What is the value of' x'?
belka [17]

Answer:

7..the way you worded the options was confusing

3 0
2 years ago
BRAINIEST ANSWER HERRREEE!!! :D
steposvetlana [31]
By asking students to answer un-biased, and with detail. Comparing and contrasting each others' genre preference is also useful.

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