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crimeas [40]
3 years ago
6

What is the domain of the function f(x) = 3|x + 4| + 1?

Mathematics
2 answers:
Gwar [14]3 years ago
8 0
The domain is all real numbers.
forsale [732]3 years ago
5 0
<h3>A. allow real numbers </h3>

I have nothing else to say

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I’m confused 273 - 75 = 280 - 82
Sati [7]

Answer: Both answers equal 198

Step-by-step explanation:

While at first this problem may look weird and even stump you. However, break it up.

What is 273-75? 198

What is 280-82? 198

The reason both equation equals each other are because both have the same answer.

Another example problem: 100-1=26+73

100-1= 26+73

100-1 = 99

26+73 = 99

5 0
3 years ago
Read 2 more answers
This your time brubru 2+2=4-1=x<br> what is x?
Ira Lisetskai [31]

the answer is three look where it says 4-1 is 3

6 0
4 years ago
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Solve the system of equations: x – 4y = –8 and –3x + 12y = 24.
ale4655 [162]
It should be C
Because when u multiply ( -3)*(x-4y= -8)

It’ll equal to -3x+12y=24 which is the same equation to the other one.
When the equations r the same, it means everything on the line is the solution for this system of equation.
4 0
3 years ago
Read 2 more answers
Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s
stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.  

Then we have n_{1} = 25, \bar{x}_{1} = 2.04, \sigma_{1} = 0.010 and n_{2} = 20, \bar{x}_{2} = 2.07, \sigma_{2} = 0.015. The pooled estimate is given by  

\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative).  

The test statistic is T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}} and the observed value is t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257. T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value t_{0} falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by 2P(T0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}, i.e.,

-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}

where t_{0.025} is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

-0.03\pm(2.0167)(0.012459)(0.3), i.e.,

(-0.0225, -0.0375)

8 0
3 years ago
A bag contains 3 red marbles, 6 blue marbles and 2 green marbles. If two marbles are drawn out of the bag, what is the probabili
Ymorist [56]
4/5 is the probability.
8 0
3 years ago
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