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3 years ago

Hi :) anyone able to teach this? I don’t understand the question , thanks in advance!

Mathematics

1 answer:

puteri [66]3 years ago

7 0

a) 62.855m

We will need to use the Pythagorean Theorem here, as we know a and c, but not b. I have attached an image of a triangle sketch to set up the problem, which should hopefully help us to visualize this problem a bit better.

Pythagorean Theorem: a^2 + b^2 = c^2

(19.8)^2 + b^2 = (65.9)^2

392.04 + b^2 = 4342.81

b^2 = 3950.77

b = 62.855 (rounded to 3 places)

b) m = 0.315

The gradient is also known as the slope. I've shown what the points would be in the image attached. Now that we know the value of b (x in the points in image), we can use points E and C to find the slope.

Point 1: (0,0)

Point 2: (62.855, 19.8)

m = (19.8 - 0) / (62.855 - 0)

m = 19.8 / 62.855

m = 0.315 (rounded to three places)

c) 17.485 degrees

The angle of inclination would be angle C. To find angle C, we will need to use an inverse trigonometry function. Any can be used since we know all of the side lengths, but I will show Sine here, opposite / hypotenuse.

sin(x) = 19.8 / 65.9

x = sin^-1 (19.8/65.9)

x = 17.485 degrees (rounded to 3 places)

Hope this helps!! :)

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Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

Step-by-step explanation:

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".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

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 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by:

$z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811$

The confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

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Answer:

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Step-by-step explanation:

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x^2 + 4 x = 4

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Subtract 2 from both sides:

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Subtract 2 from both sides:

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In the diagram below, which distances would need to be the same in order for point A to be on the bisector of angle W? Explain w

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