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marissa [1.9K]
3 years ago
11

How can you check to see if two ratios form a proportion? Explain the method you used to find the answer to the previous problem

.
Mathematics
2 answers:
antiseptic1488 [7]3 years ago
6 0

Answer:

In problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. To find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

Step-by-step explanation:

In problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. To find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

Sedaia [141]3 years ago
3 0

Answer:

Write a proportion using the ratios.

Step-by-step explanation:

Find the cross products. If the cross products are equal, then the ratios form a proportion.

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Rotate ΔABC with vertices A(-1, 4), B(2, 1), and C(3, -3) about the origin by 90°.
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Jamie is conducting inventory at a public library, so they have to verify that each item from the catalog is in its correct loca
Andrei [34K]

Answer:

Mean: 33.3

SD: 32.8

Step-by-step explanation:

4 0
3 years ago
12. Using the value of 0.538 for the proportion of those surveyed who said they exercise at least 30 minutes three or more days
ICE Princess25 [194]

Answer:

ME=1.96\sqrt{\frac{0.538 (1-0.538)}{15235}}=0.00792    

The margin of error on this case is approximated \pm 0.792\% who is very clos to the +1% reported by the Gallup.

Step-by-step explanation:

Assuming this previous info : "Results are based on telephone interview conducted as part of the Gallup-Healthways Well-Being Index survey June 1–30,2013, with a random sample of 15,235 adults, aged 18 and older, living in all 50 U.S. states and the District of Columbia.

For results based on the total sample of national adults, one can say with 95% confidence that the maximum margin of sampling error is ±1 percentage point."

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The estimated proportion on this case is \hat p =0.538 and n =15235, the margin of error is given by:

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)

And if we replace we got:

ME=1.96\sqrt{\frac{0.538 (1-0.538)}{15235}}=0.00792    

The margin of error on this case is approximated \pm 0.792\% who is very clos to the +1% reported by the Gallup.

5 0
3 years ago
Find the length of the curve given by ~r(t) = 1 2 cos(t 2 )~i + 1 2 sin(t 2 ) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplif
xxMikexx [17]

Answer:

The length of the curve is

L ≈ 0.59501

Step-by-step explanation:

The length of a curve on an interval a ≤ t ≤ b is given as

L = Integral from a to b of √[(x')² + (y' )² + (z')²]

Where x' = dx/dt

y' = dy/dt

z' = dz/dt

Given the function r(t) = (1/2)cos(t²)i + (1/2)sin(t²)j + (2/5)t^(5/2)

We can write

x = (1/2)cos(t²)

y = (1/2)sin(t²)

z = (2/5)t^(5/2)

x' = -tsin(t²)

y' = tcos(t²)

z' = t^(3/2)

(x')² + (y')² + (z')² = [-tsin(t²)]² + [tcos(t²)]² + [t^(3/2)]²

= t²(-sin²(t²) + cos²(t²) + 1 )

................................................

But cos²(t²) + sin²(t²) = 1

=> cos²(t²) = 1 - sin²(t²)

................................................

So, we have

(x')² + (y')² + (z')² = t²[2cos²(t²)]

√[(x')² + (y')² + (z')²] = √[2t²cos²(t²)]

= (√2)tcos(t²)

Now,

L = integral of (√2)tcos(t²) from 0 to 1

= (1/√2)sin(t²) from 0 to 1

= (1/√2)[sin(1) - sin(0)]

= (1/√2)sin(1)

≈ 0.59501

8 0
3 years ago
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