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Tems11 [23]
3 years ago
9

Various pieces of string are cut at different lengths. Stephanie selects three pieces that measure 4.5 cm, 6 cm, and 8.5 cm. Det

ermine what type of triangle Stephanie will form.

Mathematics
2 answers:
Yakvenalex [24]3 years ago
7 0

Answer:

Scalene triangle

Step-by-step explanation:

In this question, we are asked to determine what type of triangle will be formed given the length of the pieces which was used to form the triangle.

A triangle is one of the simplest shape in geometry. In fact it is one of the basic shapes that we have in geometry. It is a 3 sided polygon that contains 3 different edges or tips and 3 vertices.

According to the values in the question, we can see that the lengths of the pieces given are different. Now what does this mean for the triangle formed?

The type of triangle formed here would be a scalene triangle. A scalene triangle is a type of triangle in which all of its 3 sides have different lengths. Consequentially, the angles formed by these three sides also are different.

Lesechka [4]3 years ago
3 0

Answer:

Step-by-step explanation:

defining: a = 4.5 cm ; b = 6 cm ; c = 8.5 cm

Then by cosines law (see jpg adjunt):

Cos(A) = \frac{b^{2} +c^{2} -a^{2} }{2bc}  = \frac{6^{2} +8.5^{2} -4.5^{2} }{2*6*8}  =0.862

Apliying arcCosine(): A = 30.4°

Doing the same to B:

Cos(B) = \frac{c^{2} +a^{2} -b^{2} }{2bc}  = \frac{8.5^{2} +4.5^{2} -6^{2} }{2*4.5*8}  = 0.784

Apliying arcCosine(): B = 38.3°

Since, , B and C, has to:

A+B+C = 180° ---> C = 111.3°

This means: the triangule has two angules <90° and one >90°, wich is the definition of <em>obtuse triangle.</em>

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A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m
love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is c = 20.6153.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that \mu = 20.4, \sigma = 1.23

Consider the sample mean weight of 100 watermelons of this variety. This means that n = 100.

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when X = 20.27.

Since we are working with the sample mean, we use s instead of \sigma in the Z score formula

Z = \frac{X - \mu}{s}

Z = \frac{20.27 - 20.4}{0.123}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of X = c that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use Z = 1.75

Z = \frac{X - \mu}{s}

1.75 = \frac{c - 20.4}{0.123}

c - 20.4 = 0.123*1.75

c = 20.6153

This value is c = 20.6153.

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I need help Plzz &amp; Thank You .
Afina-wow [57]

{x50506}^{?}

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