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BaLLatris [955]
3 years ago
8

Find the distance between the points (-10,-10) and (-2,5)

Mathematics
1 answer:
Digiron [165]3 years ago
5 0

Answer:

17

Step-by-step explanation:

Use the distance formula

d=√(x2-x1)^2+(y2-y1)^2

√(-2+10)^2+(5+10)^2

√(8)^2+(15)^2

√64+225

√289

=17

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Natasha_Volkova [10]
Hi there! I can help you.

Rate: To find the rate, let’s divide interest earned by the principal. When you do, you get 0.15. Multiply by 100 to get 15 and divide by 3 to get 5. The simple interest rate is 5%.

New balance: All you have to do is add the principal and he new balance. When you do, you get 517.5. The new balance is $517.50.
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3 years ago
Parallel to y =9x 3, y-intercept at -2
Mashutka [201]
Slope-intercept form:
y=mx+b
m=slope
b=y-intercept

We have this line: y=9x+3; a line parallel to this line (y=9x+3) will have the same slope; therefore m would be equal to 9 (m=9).

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m=9
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3 years ago
What’s 11 divide by 3 and 2/3
Archy [21]

Answer:

3

Step-by-step explanation:

8 0
3 years ago
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Someone please help<br><br><br><br><br><br><br><br> give 20 points
Leokris [45]

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14 = 14

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From least-to-greatest:

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6 0
3 years ago
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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).

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3 years ago
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