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krok68 [10]
2 years ago
7

Please solve the whole page

Mathematics
1 answer:
Levart [38]2 years ago
6 0

Answer:

1. 3/20    

2. 76/4

3. 16

4. 1/8

5. 11/1

6. 2.5/1

7. 13/1

8. 255/1

9. 20, 140, 240

10. 3.5, 17.5, 31.5

11. 12

12. 2      

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A^4 + 2b OVER a; if a =3 and b = 12
Tems11 [23]

Answer:

105

Step-by-step explanation:

Given : a = 3, b = 12

3 ⁴ = 81

2 x 12 = 24

Result:

81 + 24

= 105

6 0
3 years ago
Did I get this right or no I didn’t
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7 0
3 years ago
7/8 yard of material is needed to make a blanket, how many blankets can be made from 21 yards of material?
dmitriy555 [2]

Answer:

24

Step-by-step explanation:

the question translates to asking how many times does 7/8 fit in 21, so 21 divided by 7/8.

when we divide by a fraction, we can multiply by it's inverse, so 21 divided by 7/8, is 21 multiplied by 8/7. so:

21×8/7 = (21×8)/7 = (21/7)×8 = 3×8 = 24

8 0
2 years ago
What are the two plot lines for this
Lyrx [107]

Answer:

You can plot the lines by putting the values of x and y.

For y=x-2, if we put x= 2, then we got y=0

So, the point for this equation will be (2,0). Similarly, you can get many points for the equations and join them to plot lines in the graph.

4 0
3 years ago
A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 :
const2013 [10]

Answer:

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.

This means that n = 1067, \pi = \frac{74}{1067} = 0.069

80% confidence level

So \alpha = 0.2, z is the value of Z that has a pvalue of 1 - \frac{0.2}{2} = 0.9, so Z = 1.28.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 - 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.059

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 + 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.079

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

7 0
2 years ago
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