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77julia77 [94]
3 years ago
13

Another way to write the value absolute value inequality |p|<12

Mathematics
2 answers:
mote1985 [20]3 years ago
5 0

Answer:

-12 <p <12

Step-by-step explanation:

|p|<12

We can write this without the absolute values

Take the equation with the positive value on the right hand side  and take the equation with a negative value on the right side remembering to flip the inequality.  Since this is less than we use and in between

p < 12 and p >-12

-12 <p <12

Lady bird [3.3K]3 years ago
4 0

Step-by-step explanation:

For\ a>0\\\\|x|a\Rightarrow x>a\ \wedge\ x

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Taylor and Katy repeatedly toss a fair coin. Katy wins if heads are thrown, Taylor wins if tails are thrown, and the game ends w
ehidna [41]

Answer:

25%

Step-by-step explanation:

In order to find the chance/probability that Katy wins we need to first find the probability of her winning a single coin toss and then multiply it twice. This is because Katy has 1 win already and needs two more win the entire game. Since a coin toss only has 2 options and only 1 of those outcomes will cause Katy to win, then this means that the probability of Katy winning is 1/2 or 50%. Now we can multiply this probability twice to get the probability of Katy winning the entire game

1/2 * 1/2 = 1/4 or 25%

5 0
3 years ago
Write an equation of the direct variation that includes the point (9, –12).
Korvikt [17]
Y = (-4/3)x that is the right answer
5 0
3 years ago
Read 2 more answers
1. A waste management service attempts to design routes so that each of their trucks pick-up on average four tons of garbage or
IRINA_888 [86]

Answer:

1.-Then we p-value indicates that we are in the rejection region we reject H₀

We support the claim of the garbage collector the average picking up more than 4 tn of garbage

2.- p  = 75 %

3.-3.-t(s) is in the rejection region we accept H₀ we have not evidence to support Marc´s claim

Step-by-step explanation:

1.- If p-value is 0,04   and significance level α = 5 %  or  α = 0,05 then p-value < α

Test hypothesis should be    ( x the average of garbage)      

Null hypothesis          H₀           x = 4 Tn

Alternative Hypothesis     Hₐ    x  > 4 Tn

Then  alternative hypothesis suggests a one tail-test to the right  and

p-value <  0,05

Then we p-value indicates that we are in the rejection region we reject H₀

We support the claim of the garbage collector the average picking up more than 4 tn of garbage

2.- As we are dealing with a normal distribution the CI  95 % is symmetrical with respect to the mean, therefore the proportion of student living in Brooklyn is:  

(0,73 + 0,77) /2

p = 0,75     p  = 75 %

Test hypothesis:

Null Hypothesis            H₀            μ    =  26

Alternative Hypothesis      Hₐ      μ   <  26

Alternative hypothesis tells us the test is a one-tail test to the left

Sample size   n = 25

Sample mean   μ  =  24,4

Sample Standard deviation  =  9,2

We assume normal distribution, and as n < 30 we use t-student table

with  24 degree of freedom

Significance level is 0,05  and df = 24  we find t (c)  in t- student table

t(c) = 1,7109      test to the left   t(c) = -1,7109

To calculate  t(s)

t(s)  =  (  24,4 - 26 ) / 9,2 / √25

t(s) =  - 1,6 * 5 / 9,2

t(s) = - 0,87

Comparing t(s)  and t(c) we have

t(s) > t(c)

t(s) is in the rejection region we accept H₀ we have not evidence to support Marc´s claim

6 0
3 years ago
Draw an area model. Then, Solve using the standard algorithm.<br> 642 x 257
Darina [25.2K]

Answer:

642 × 257 = 164,994

I hope it is helpful

4 0
3 years ago
Please help me with #'s 5&amp;6
Fynjy0 [20]

5.\\(x+3)^2+7=-2\qquad|-7\\\\(x+3)^2=-9 < 0\\\\\text{NO REAL SOLUTION}\\\\(x+3)^2=-9\to x+3=\pm\sqrt{-9}\\\\x+3=-3i\ \vee\ x+3=3i\qquad|-3\\\\\boxed{x=-3-3i\ \vee\ x=-3+3i}

6.\\4(x-10)^2=25\qquad|:4\\\\(x-10)^2=\dfrac{25}{4}\to x-10=\pm\sqrt{\dfrac{25}{4}}\\\\x-10=-\dfrac{\sqrt{25}}{\sqrt4}\ \vee\ x-10=\dfrac{\sqrt{25}}{\sqrt4}\\\\x-10=-\dfrac{5}{2}\ \vee\ x-10=\dfrac{5}{2}\\\\x-10=-2.5\ \vee\ x-10=2.5\qquad|+10\\\\\boxed{x=7.5\ \vee\ x=12.5}

7 0
3 years ago
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