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2 years ago

Can the triangles below be proven similar?

Mathematics

1 answer:

jeka942 years ago

3 0

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Yes ! they ofcourse are similar and here goes the explanation behind that ~

two triangles are said to be similar by SAS postulate in case the ratios of their side lengths comes out to be equal along with one angle lying in between the side lengths

in this case ,

$\frac{AB}{BD} \: should \: be \: equal \: to \: \frac{EB}{BC} \\$

so , let's check if the ratio is equal or no ~

$\implies\frac{AB}{BD} = \frac{27}{18} = \frac{3}{2} \\ \\ \implies \: \frac{EB}{BC} = \frac{36}{24} = \frac{3}{2} \\ \\ \therefore \: \frac{AB}{BD} = \frac{EB}{BC}$

Also ,

∠ABE = ∠CBD

since they both are vertically opposite , i.e. , angles with the same vertex B .

thus the triangles are similar by SAS postulate !

hence , Option A is correct !

hope helpful ~

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Genrish500 [490]

Answer:

a) 0.057

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c) 0.4766

Step-by-step explanation:

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

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$z=\frac{185-175}{20/\sqrt {10}}=1.5811$

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As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of 1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of 1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is

$\boxed {p=0.057}$

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute $\bar x_{critical}$