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Art [367]
1 year ago
12

Find the range of the function f(x) = 3x² - 2x for the domain (1, 2, 3).

Mathematics
1 answer:
TiliK225 [7]1 year ago
4 0

Answer:

Step-by-step explanation:

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paula spent 3/8 of her allowance on clothes and 1/6 on entertainment. what fraction of her allowance did she spend on other item
MArishka [77]
We have to calculate the fraction of Paula`s allowance that she spent on other items if she already had spent 3/8 on clothes and 1/6 on entertainment. First we have to add: 3 / 8 + 1 / 6 = ( LCD is 24 ) = 9 / 24 + 4 / 24 = 13 / 24. Then : 1 - 13 / 24 = 24 / 24 - 13 / 24 = 11 / 24. Answer: She has spent 11 / 24<span> of her allowance on other items. </span>
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3 years ago
Tell what number you would add to each side of the inequality. please answer ASAP
Novay_Z [31]

Step-by-step explanation:

x - 12 < 15

x < 15 + 12

x < 27

For x < 27 , i will add a number smaller than 27.

x = {26,25,24,23, etc}

5 0
3 years ago
Find the value of x. WILL GIVE BRAINLIEST!!!
Readme [11.4K]

Answer: x = 16

Step-by-step explanation:

24 + 2 + 4x = 90

26 + 4x = 90

4x = 64

x = 16

6 0
3 years ago
(1,10), (18.-10) slope
kobusy [5.1K]

Answer:

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8 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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