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BabaBlast [244]
3 years ago
6

Given: △ABC; AB=BC, m∠BDA = 60°, BD=4 cm, BD ⊥ BA . Find: DC, AC.

Mathematics
1 answer:
vekshin13 years ago
8 0

Answer:

DC = 10.93 cm ,  AC = 9.8 cm

Step-by-step explanation:

From trigonometry;

⇒ Tan 60 = AB/BD

⇒AB = BD Tan 60 ( where BD = 4 cm )

⇒ AB = 6.93 cm

Also, AB=BC , therefore;

⇒ BC = 6.93 cm

⇒ Cos 60 = BD/AD

⇒ AD = BD/ Cos 60 = 4/Cos 60

⇒ AD = 8 cm

From Pythagoras theorem;

⇒ AC^{2} = AB^{2} + BC^{2} = (6.93)^{2} + (6.93)^{2}

⇒ AC = \sqrt{96.05} = 9.80 cm

⇒ DC = BD + BC = 4 + 6.93

⇒ DC = 10.93 cm

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3. The average age of online consumers a few years ago was 23.3 years with a standard deviation of 5.3 years. It is believed the
Slav-nsk [51]

Answer:

We cant say, with α =0.01, that the average age changed. The p-value is 0.014

Step-by-step explanation:

As a consecuence of the Central Limit Theorem, the mean sample has a distribution approximately normal, with unknown mean and standard deviation \sigma = 5.3/\sqrt{20} = 1.1851 (the standard deviation of one single sample divided by the  sqaure root of the sample lenght).

The null hypothesis H₀ is that the average age is still 23.3 (the mean is 23.3). The alternative hypothesis is that the mean is different. We want to see if we can refute H₀ with significance level α =0.01.

Lets call X the mean of a random sample of 20 online shoppers. As we discuse above, X is approximately normal with unknown mean and standard deviation equal to 1.1851 . If we take the hypothesis H₀ as True, then the mean will be 23.3, and if we standarize X, we have that the distribution

W = \frac{X - 23.3}{1.1851}

Will have a distribution approximately normal, with mean 0 and standard deviation 1.

The values of the cummulative ditribution of the normal function can be found in the attached file. Since we want a significance level of 0.01, then we need a value Z such that

P(-Z < W < Z) = 0.99

For the symmetry of the standard Normal distribution, we have that Φ(Z) = 1-Φ(-Z), where Φ is the cummulative distribution function of the standard Normal random variable. Therefore, we want Z such that

Φ(Z) = 0.995

If we look at the table, we will found that Z = 2.57, thus,

0.99 = P(-2.57 < W < 2.57) = P(-2.57 < \frac{X-23.3}{1.1851} < 2.57)\\ = P(-2.57*1.1851 + 23.3 < X < 2.57*1.1851 + 23.3) = P(20.254 < X < 26.345)

Thus, we will refute the hypothesis if the observed value of X lies outside the interval [20.254, 26.345].

Since the observed value is 26.2, then we dont refute H₀, So we dont accept that the number average age of online consumers has changed.

For us to refute the hypothesis we need Z such that, for the observed value, |W| > Z. Replacing X by 26.2, we have that [tex] W = \frac{26.2-23.3}{1.1851} = 2.4470. We can observe that Φ(2.447) = 0.993, substracting that amount from 1 and multiplying by 2 (because it can take low values too), we obtain that the p-value is 0.014.  

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8 0
3 years ago
From midnight to 7:00 am, the temperature dropped 0.8°C
sp2606 [1]

Answer:

5.2

Step-by-step explanation:

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1 year ago
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Answer:

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7 0
2 years ago
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Need help tremendously
Alina [70]
<span>basically since 30 minutes is half of an hour, to get to a full hour you can multiply by 2, so you do that for both, 2/3 included so you get 2/3 * 2/1 4/3 is 1 1/3</span>
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3 years ago
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kati45 [8]

Answer:

<em>NO</em>

Step-by-step explanation:

They are equal

83 tenths divided by 10 is 8.3

<u>Hope this helps :-)</u>

8 0
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