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3 years ago

HEEEEEEEEEEEELP HEEEEEEELP

Mathematics

1 answer:

Bingel [31]3 years ago

4 0

Answer:

x < 2

Step-by-step explanation:

We're going to solve this just like we would an equation, but with just one little twist. First, let's add 1/3 to both sides.

5/6x - 1/3 > 1 1/3

5/6x > 1 2/3

First, let's turn 1 2/3 into an improper fraction to make it easier to work with. Multiply the big number by the denominator, then add the numerator.

1 x 3 + 2 = 5

1 2/3 = 5/3

Now, multiply 5/3 by 2/2 to get 6 as the denominator. This works because 2/2 simplifies to 1 and therefore does not change the value of the fraction.

5 x 2 = 10

3 x 2 = 6

5/3 = 10/6

Now divide both sides by 5/6. Because they both have the same denominator, we can cancel the denominator out and just divide 5x/5 and 10/5.

5/6x > 10/6

5x > 10

x > 2

Wait! We don't have the correct answer just yet. Whenever you divide an inequality, you have to flip the inequality sign.

x < 2

Now <em>there's</em> our final answer.

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Answer:

a) Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

b) $t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

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Step-by-step explanation:

Information given

$\bar X=130$ represent the sample mean for the amount spent each shopper

$s=40$ represent the sample standard deviation

$n=80$ sample size

$\mu_o =120$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value f

Part a

We want to verify if the shoppers participating in the loyalty program spent more on average than typical shoppers, the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

The statistic for this case would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$

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3 years ago

Read 2 more answers

WILL MARK BRAINIEST FOR FIRST PERSON TO ANSWER WITH THE STEPS

kondaur [170]

Answer:

a. 1.2 ohms

b. 6 ohms

Step-by-step explanation:

As with any problem involving evaluation of a formula, substitute the given values for the corresponding variables and do the arithmetic.

$R=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}}=\dfrac{1}{\dfrac{1}{2}+\dfrac{1}{3}}=\dfrac{1}{\left(\dfrac{5}{6}\right)}=\boxed{\dfrac{6}{5}=1.2}$

$R=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}}=\dfrac{1}{\dfrac{1}{10}+\dfrac{1}{15}}=\dfrac{1}{\left(\dfrac{5}{30}\right)}=\dfrac{30}{5}=\boxed{6}$

_____

<em>Comment on these problems</em>

You will notice that the values in part (b) are 5 times those in part (a). So, it is no surprise that the answer in part (b) is 5 times the answer in part (a).

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