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Elina [12.6K]
3 years ago
11

An example of Power of a Power Property

Mathematics
1 answer:
ella [17]3 years ago
3 0
Power of a power as applied to exponents is the following:

The power of a power is the product of the exponents, or

(x^a)^b = x^(ab)

For example
(5^2)^3=5^(2*3)=5^6=15625
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Solve the system of equations<br> −4x+3y=−2 <br> y=x−1 ​ <br><br> x = <br> y =
dolphi86 [110]
Answer: x = -1 and y = -2

Step-by-step explanation:

So if you know y = x-1 you can plug that in where y is in the first equation

-4x + 3(x-1) = -2

Distribute the 3:

-4x +3x -3 = -2

Combine like terms:
-x -3 = -2

Add 3 to both sides:
-x = 1

Multiply by -1:
X = -1

Then you plug that into the other equation to get y=(-1)-1

Which is -2
4 0
3 years ago
Keaton's dinner bill at a local restaurant before tax and tip was $25.85 if he wants to leave 15% of the bill for tip and there'
Ksju [112]
Bill = $25.85

Tips = 15% x $25.85 = $3.88

Tax = 6.25% x $25.85 = $1.62

Total amount = $25.85 + $3.88 + $1.62 = $31.35

------------------------------------------------------
Answer: He will spend $31.35 in total.
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8 0
4 years ago
MATH HELP!!!!!<br><br> If y = 15 when x = 2, what is the value of x when y = 7?
Ede4ka [16]

<u>Direct Variation:</u>  \frac{y}{x} = k

\frac{15}{2} = k

when y = 7:      \frac{7}{x} = \frac{15}{2}

14 = 15x

\frac{14}{15} = x

<u>Inverse variation:</u>   y*x = k

15 * 2 = k

30 = k

when y = 7:      7 * x = 30

x = \frac{30}{7}


3 0
3 years ago
Which ones do I choose
weeeeeb [17]
It should be B because is a minus is in front of something that is a negative they change to be positive

7 0
3 years ago
To compare two programs for training industrial workers to perform a skilled job, 20 workers are included in an experiment. Of t
nikklg [1K]

Answer:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

p_v =P(t_{(18)}

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

Step-by-step explanation:

Data given and notation

We can calculate the sample mean and deviation with these formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X_{1}=19.1 represent the mean for the sample mean for 1

\bar X_{2}=23.3 represent the mean for the sample mean for 2

s_{1}=4.818 represent the sample standard deviation for the sample 1

s_{2}=5.559 represent the sample standard deviation for the sample 2

n_{1}=10 sample size selected 1

n_{2}=10 sample size selected 2

\alpha represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the average time taken when training under method 1 is less than the average time for Method 2, the system of hypothesis would be:

Null hypothesis:\mu_{1} \geq \mu_{2}

Alternative hypothesis:\mu_{1} < \mu_{2}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

P-value

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=10+10-2=18

Since is a one sided test the p value would be:

p_v =P(t_{(18)}

Conclusion

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

4 0
3 years ago
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