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dolphi86 [110]
3 years ago
9

Find the distance between the points (6, 5 square root 2) and (4, 3square root 2).

Mathematics
2 answers:
Bond [772]3 years ago
7 0

For this case we use the distance formula between two points.

We have then:

d=\sqrt{(x2-x1)^2+(y2-y1)^2}

Substituting values in the given equation we have:

d=\sqrt{(6-4)^2+(5\sqrt{2}-3\sqrt{2})^2}

Rewriting the equation we have:

d=\sqrt{(2)^2+(2\sqrt{2})^2}

d=\sqrt{4+8}

d=\sqrt{12}

d=\sqrt{3*4}

d=2\sqrt{3}

Answer:

The distance between the points is:

d=2\sqrt{3}

Sonbull [250]3 years ago
4 0

The distance would be 2√3 , in other words it is 2 square root 3

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Step-by-step explanation:

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Answer:

We conclude that the mean IQ of her students is different from the national average.

Step-by-step explanation:

We are given that the national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15.

The dean of a college want to test whether the mean IQ of her students is different from the national average. For this, she administers IQ tests to her 144 students and calculates a mean score of 113

Let, Null Hypothesis, H_0 : \mu = 100 {means that the mean IQ of her students is same as of national average}

Alternate Hypothesis, H_1 : \mu\neq 100  {means that the mean IQ of her students is different from the national average}

(a) The test statistics that will be used here is One sample z-test statistics;

               T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ N(0,1)

where, Xbar = sample mean score = 113

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So, test statistics = \frac{113-100}{\frac{15}{\sqrt{144} } }

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Now, at 0.05 significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means our test statistics will fall in the rejection region and we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean IQ of her students is different from the national average.

3 0
3 years ago
Need help with this dont understand it at all
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