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Cerrena [4.2K]
3 years ago
12

In an isosceles triangle, the perimeter is 75 cm and one of the sides is 25 cm. Find all its sides. Can you find all angles of t

he triangle? Explain your answer.
Mathematics
2 answers:
ioda3 years ago
6 0
All are 25 cm is the answer

kramer3 years ago
4 0
A triangle has 3 sides.

To find the perimeter, you would take the length off all sides and add them together.

Since the perimeter of the triangle is 75 cm, and one side is 25, you would take the perimeter and divide it by the number of sides in the triangle.

75/3=25

To check your work you would take 25 and multiply that by 3. If you get 75, then you got it right ;-)

Therefore, 25cm would be your answer for all the sides of the triangle.

-

In order to find the angles of all the sides, a triangle has a total of 180°.

You would then take 180° and divide it by the number of sides the triangle has which is 3.

180°/3= 60°

You would check your work by multiplying 60 by 3 which is 180

60° would then be the angles of all the sides in the triangle.

Hope this helps explain!
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In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi
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Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n  = 50

sample mean \overline x = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

X \sim N ( \mu \sigma)

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})

P(X>140) = 1 - P( Z\leq0.42)

From z tables,

P(X>140) = 1 - 0.6628

P(X>140) = 0.3372

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

Y \sim Binomial (np)

Y \sim Binomial (50,0.3372)

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49}  + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +...  + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}

P(Y ≥ 15) = 0.763

7 0
3 years ago
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