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Alex
3 years ago
8

Given the data

Mathematics
1 answer:
prisoha [69]3 years ago
3 0
I dnshdjjdndndndjiehdkxkuwhdndnxjmd
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3×1 3/4 greater than less than or equal to 1 3/4
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3•1 3/4 is Greater than 1 3/4
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3 years ago
John is using two right triangles to build a rabbit cage. Can his right triangles also be isosceles? Explain
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Sure, there's such a thing as an isosceles right triangle.   It's what you get when you draw the diagonal of a square.  It has one right angle (of course) and two 45 degree angles.   It's the shape that vexed the Pythagoreans.

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3 years ago
The product of a number and 7 is less than 28.
Debora [2.8K]

Answer:

3.9 or less

Step-by-step explanation:

This is because 4 times 7 is equal to 28, so it has to be less than that!

4 0
3 years ago
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The
Tems11 [23]

Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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 z-score that has a p-value of 1 - \frac{\alpha}{2}.

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that n = 387, \pi = \frac{298}{387} = 0.77

84% confidence level

So \alpha = 0.16, z is the value of Z that has a p-value of 1 - \frac{0.16}{2} = 0.92, so Z = 1.405.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

6 0
3 years ago
What’s the answer to this?
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Y=6 x is 3 and y is 6 the answer is 6 from the y axis
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2 years ago
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