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frez [133]
3 years ago
15

If n represents a number,then write an expression for three times the difference of the number and six increased by four times t

he number
Mathematics
1 answer:
Contact [7]3 years ago
3 0

Write and simplify the expression:

3(n-6)+4n

3n-18+4n

7n-18

Solution: 7n-18

I hope I helped you with your solution! If you could give me brainliest, I would be very thankful!

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What is 65 5/6+ 6 4/6 in simplest form?
juin [17]
72.5 as decimal or 72 1/2 as fraction
8 0
3 years ago
Use the greatest common factor to write the fraction in simplest form: 32/48
Illusion [34]

Answer:

8

Divide both by 8

Which gives you

4/6

That can be simplest form by 2

2/3 is your final answer!

8 0
3 years ago
Need help with this question
goldfiish [28.3K]

Answer:

x = 12

Step-by-step explanation:

Complementary angles add up to have a sum of 90 degrees. We can create an equation and solve for x.

3x + 6x - 18 = 90

~Simplify

9x - 18 = 90

~Add 18 to both sides

9x = 108

~Divide 9 to both sides

x = 12

Best of Luck!

4 0
3 years ago
What is the product ?
Olenka [21]

Answer:

The answer to your question is:

Step-by-step explanation:

   \frac{x^{2}-16 }{2x + 4} \frac{x^{3}-2x^{2} + x }{x^{2}+ 3x - 4}

Factorize    \frac{(x+4)(x-4)}{2(x+2)}  \frac{x(x^{2}-2x + 1) }{(x+4)(x-1)}

Factorize    \frac{(x+4)(x-4)}{2(x+2)} \frac{x(x-1)^{2} }{(x+4)(x-1)}

Simplify      \frac{x(x-4)}{2(x+2)}

7 0
3 years ago
A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's p
Grace [21]

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, H_0 : p \leq 0.29  {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, H_1 : p > 0.29  {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

          T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = proportion of chips that do not fail in the first 1000 hours of their use = 32%

            n = sample of chips = 1500

So, <u>test statistics</u> = \frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }

                              = 2.491

<em>Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.</em>

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

7 0
3 years ago
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