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kow [346]
3 years ago
8

Can someone solve these please using the formula AB | a - b | And can you show work

Mathematics
1 answer:
Lana71 [14]3 years ago
4 0

Answer: 1) AC = 13

The formua does not actually apply to all of the problems.

Step-by-step explanation:

1) The absolute value of -8 is added to the absolute value of 5. 8+5=13

2) Subtract the length of the EF from EG to get the length of FG 21 -6 = 15

3) Take what's given and create an equation to solve. 4x + 15 +39 =110. 4x = 110 - (15+39).

4x =110-54. 4x =56. x=56/4. x=14

4) Create another equation. You have two segments that add up to the length of EG, given =23

EF+FG=EG

(2x-12)+(3x-15)=23

5x - 27 = 23

5x= 23+27 5x =50. x = 10

Substitute 10 for x

EF=2(10) -12 EF=8

FG=3(10)-15. FG=15

EF+FG =EG.

8 + 15 = 23

5) 2/5 of 25 is 10 So EF is 10. Subtract from 25 to get FG

FG = 15

I hope this helps you.

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A researcher at the University of Washington medical school believes that energy drink consumption may increase heart rate. Supp
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Answer:

Step-by-step explanation:

Given that:

mean μ = 70

sample size = 25

sample mean = 73

standard deviation = 7

level of significance = 0.10

The null hypothesis and the alternative hypothesis can be computed as follows:

\mathtt{H_o : \mu = 70} \\ \\  \mathtt{H_1 : \mu > 70 }

The z score for this statistics can be calculated by using the formula:

z = \dfrac{X- \mu}{\dfrac{\sigma}{\sqrt{n}}}

z = \dfrac{73- 70}{\dfrac{7}{\sqrt{25}}}

z = \dfrac{3}{\dfrac{7}{5}}

z = \dfrac{3 \times 5}{{7}{}}

z = 2.143

At level of significance of 0.10

degree of freedom = n -1

degree of freedom = 25 - 1

degree of freedom = 24

The p - value from the z score at   level of significance of 0.10 and degree of freedom of 24 is:

P - value = 1 - (Z < 2.143)

P - value =  1 - 0.9839

P - value =  0.0161

Decision Rule: since P value is lesser than the level of significance, we reject the null hypothesis.

Conclusion: We conclude that energy drink consumption increases heart rate.

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We need to write an equation for the description.

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