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3 years ago

Help Please I'm Out Of Time ;(

Mathematics

1 answer:

trasher [3.6K]3 years ago

6 0

Answer:

Total area = 31.5 unit²

Step-by-step explanation:

Assume 1 block = 1 unit²

Divide diagram into 3 part

1. 5 by 5 block square

2. 2 by 2 block triangle(1)

2. 3 by 3 block triangle(2)

Area of square = side × side

Area of square = 5 × 5

Area of square = 25 unit²

Area of triangle(1) = [1/2][base][height]

Area of triangle(1) = [1/2][2][2]

Area of triangle(1) = 2 unit²

Area of triangle(2) = [1/2][base][height]

Area of triangle(2) = [1/2][3][3]

Area of triangle(2) = 4.5 unit²

Total area = Area of square + Area of triangle(1) +Area of triangle(2)

Total area = 25 unit² + 2 unit² + 4.5 unit²

Total area = 31.5 unit²

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3 years ago

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Step-by-step explanation:

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3 years ago

The numbers 2, 3, 4, and 5 are indicated on cards that are placed in a hat.

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3 years ago

Which transformation or sequence of transformations would produce an image that is not congruent to its pre‐image?

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2 years ago

Read 2 more answers

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

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3 years ago