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irina [24]
3 years ago
10

PLZ HELP IM 100% LOST ON THIS

Mathematics
1 answer:
xenn [34]3 years ago
8 0

The equation f(x)=g(x) is

|x-4|-11=-\sqrt{5x}

Some observations:

  • \sqrt{5x} is defined only as long as 5x\ge0, or x\ge0
  • wherever \sqrt{5x} is defined, its value must be non-negative, so that -\sqrt{5x} is never positive
  • by the definition of absolute value, we have |x-4|=x-4 if x\ge4, and |x-4|=-(x-4)=4-x if x. Then

|x-4|-11=\begin{cases}x-15&\text{for }x\ge4\\-x-7&\text{for }x

If x, the equation becomes

-x-7=-\sqrt{5x}\implies x+7=\sqrt{5x}

Taking the square of both sides gives

(x+7)^2=\left(\sqrt{5x}\right)^2\implies x^2+14x+49=5x\implies x^2+9x+49=0

but since the discriminant is 9^2-4\cdot1\cdot49, there are no real solutions.

If x\ge4, then

x-15=-\sqrt{5x}

Taking squares gives

(x-15)^2=\left(-\sqrt{5x}\right)^2\implies x^2-30x+225=5x

and solving by the quadratic formula gives two potential solutions,

x=\dfrac{35\pm5\sqrt{13}}2

which have approximate values of 8.49 and 26.51.

We know for any value of x that g(x)\le0. We have f(8.49)\approx-6.51 and f(26.51)\approx11.51, so only the first solution 8.49 is valid.

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Secants – 1. <CBG 2. <AGF 3. <ABD

Tangent – 1. <CDE

Chords – 1. <BD 2. <DF 3. <BG 4. <BF 5. <GD 6. <FG

Angles – 1. 45 2. 75 3. 35 4. 70 5. 75 6. 55 7. 50 8. 25 9. 35 10. 70 11. 50 12. 25 13. 70 14. 50. 15. 60 16. 85 17. 95 18. 85 19. 95

Step-by-step explanation:

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What percent of 68 is 41?
EastWind [94]

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The author drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 199 times. Here are the observed
Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28​, 29​, 47​, 40​, 22​, 33.

Probability of an Event

      =\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911

Result is not significant.

3.

(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378

Result is not significant.

4.

(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349

Result is not significant.

5.

(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773

Result is not significant.

6.

(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair​ die.

   

8 0
3 years ago
Carolyn answered 23 questions correctly for a score of 92%. How many questions were on the test?
Alla [95]

Answer:

25*0.92= 23 so she answered 23 question from 25 questions

Step-by-step explanation:

6 0
2 years ago
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