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Olenka [21]
3 years ago
11

A picture measuring 2.5" high by 6" wide is to be enlarged so that the width is now 18 inches. How tall will the picture be?

Mathematics
2 answers:
Usimov [2.4K]3 years ago
6 0
7.5" if the width is now 18 that means its value increased by 3 so multiply 2.5 by 3 and its = to 7.5
Law Incorporation [45]3 years ago
6 0
Since the width went from 6 to 18, we can conclude that it increased by a factor of 3, 18/6=3. We can now factor in 3 to 2.5 and we see that 3×2.5=7.5. The new height is 7.5".
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A rectangle Has an area of 252 yd.². The length of the rectangle is 10 yards longer than twice it’s worth. Find the width and le
Rama09 [41]

Answer:

Width = 9 yds

Length = 28 yds

Step-by-step explanation:

Width = x

Length = 2x + 10

Area is 252 yd²

x(2x + 10)  = 252

2x² + 10x - 252 = 0

2 (x + 14) (x - 9) = 0

(x + 14) (x - 9) = 0

x = - 14 or x = 9

Width = 9 yds

Length = 2x9 + 10 = 28 yds

7 0
3 years ago
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Zeros: x=1 (multiplicity 2) and x=-4
BARSIC [14]
Starting with x=1, subtract 1 from each side,
(x-1)=0

Our polynomial will have multiplicity of 2 for this particular zero,
(x-1)(x-1)

and do similar with the x=-4, add 4 to each side,
(x+4)=0

So our final result is: (x-1)(x-1)(x+4)

Expand out the brackets if you need this in standard form.
5 0
3 years ago
What is the range of y=sec^-1(x)?
Bingel [31]
Assuming that the x is not part of the ^-1 the range would be 

[0,π/2) U (π/2,<span>π]</span>
3 0
2 years ago
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Please help... #18 and #28
Vesna [10]
I hope this helps you

5 0
3 years ago
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Evelyn wants to estimate the percentage of people who own a tablet computer she surveys 150 indvidals and finds that 120 own a t
igomit [66]

Answer:

The 99% confidence interval for the percentage of people who own a tablet computer is between 71.59% and 88.41%

Step-by-step explanation:

Confidence interval for the proportion of people who own a tablet:

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}.

For this problem, we have that:

n = 150, \pi = \frac{120}{150} = 0.8

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.576.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 - 2.575\sqrt{\frac{0.8*0.2}{150}} = 0.7159

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 + 2.575\sqrt{\frac{0.8*0.2}{150}} = 0.8841

Percentage:

Multiply the proportion by 100.

0.7159*100 = 71.59%

0.8841*100 = 88.41%

The 99% confidence interval for the percentage of people who own a tablet computer is between 71.59% and 88.41%

8 0
3 years ago
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