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Elza [17]
3 years ago
12

In a diagram, the area of the large square is 1 square unit. two diagonal segments divide the square into four equal-sized trian

gles. Two of these triangles are divided into smaller red and blue triangles that all have the same height and base length. Find the area of the red triangle.

Mathematics
1 answer:
serg [7]3 years ago
4 0

Answer:

1/8 square unit or 0.125 square unit

Step-by-step explanation:

Please kindly check the attached files for explanation and the picture or diagram of the square with triangles

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10 and 11

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Round 5 358 708 to the nearest million​
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Traveling 1,200 miles in 4 hours
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devin makes a 15% commission each month for everything he sells. if devin sells $2750 of merchandise this month, how much of a c
kakasveta [241]

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$412.50

Step-by-step explanation:

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7 0
3 years ago
A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears
marta [7]

Answer:

(a) The probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b) <em>n</em> = 28.09

Step-by-step explanation:

The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.

It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of <em>Y</em>ₙ are:

\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}

(a)

For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\

Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:

P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z>  4.47)\\=1-P(Z

**Use a <em>z </em>table for probability.

Thus, the probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b)

It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

P(Y_{n}>1000)=0.99\\1-P(Y_{n}

The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.

Compute the value of <em>n</em> as follows:

z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n}  \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0

The last equation is a quadratic equation.

The roots of a quadratic equation are:

n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of <em>n</em> = 28.09.

8 0
3 years ago
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