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3 years ago

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For what value of x is the area of the rectangle greater than the perimeter? The base of the perimeter is x+2

1 answer:

TiliK225 [7]3 years ago

8 0

If we call the area A, then $A=7(x+2)=7x+14$
if we call the perimeter P, then $P=7*2+2(x+2)=2x+18$
$A>P$
$7x+14>2x+18$
subtracting 14 gives $7x>2x+4$
subtracting 2x gives $5x>4$
dividing by 5 gives $x>\frac{4}{5}$
or $x>0.8$

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Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl

amm1812

Answer:

1. $P(s_1|I)=\frac{1}{11}$

2. $P(s_2|I)=\frac{8}{11}$

3. $P(s_3|I)=\frac{2}{11}$

Step-by-step explanation:

Given information:

$P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3$

$P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1$

(1)

We need to find the value of P(s₁|I).

$P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}$

$P(s_1|I)=\frac{0.015}{0.165}$

$P(s_1|I)=\frac{1}{11}$

Therefore the value of P(s₁|I) is $\frac{1}{11}$ .

(2)

We need to find the value of P(s₂|I).

$P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}$

$P(s_2|I)=\frac{0.12}{0.165}$

$P(s_2|I)=\frac{8}{11}$

Therefore the value of P(s₂|I) is $\frac{8}{11}$ .

(3)

We need to find the value of P(s₃|I).

$P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}$

$P(s_3|I)=\frac{0.03}{0.165}$

$P(s_3|I)=\frac{2}{11}$

Therefore the value of P(s₃|I) is $\frac{2}{11}$ .

4 0

3 years ago

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lakkis [162]

Answer:

< W ≅ < P.

Step-by-step explanation:

< W ≅ < P.

< W is the refection of < P over the x-axis.

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3 years ago

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alisha [4.7K]

Answer:

2nd choice

Step-by-step explanation:

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3 years ago

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If you know that a < b, and both a and b are positive numbers, then what must be true about the relationship between the oppo

nlexa [21]

What ever number A is, it is going to be less than whatever number B is.

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The fourth term of an Arithmetic Sequence is equal to 3 times the first term, and the seventh term exceeds twice the third term

11Alexandr11 [23.1K]

Answer:

The first term is 3. The common difference is 2.

Step-by-step explanation:

The first term is x.

The common difference is d.

The second term is x + d.

3rd term: x + 2d

4th term: x + 3d

7th term: x + 6d

"The fourth term of an Arithmetic Sequence is equal to 3 times the first term"

x + 3d = 3 * x Eq. 1

"the seventh term exceeds twice the third term by 1"

x + 6d = 2(x + 2d) + 1 Eq. 2

Simplify Eq. 1:

2x = 3d

Simplify Eq. 2:

x + 6d = 2x + 4d + 1

x = 2d - 1

Multiply both sides of the last equation by 2.

2x = 4d - 2

2x = 3d (simplified Eq. 1)

Since 2x = 2x, then the right sides are equal.

3d = 4d - 2

d = 2

2x = 3d

2x = 3(2)

2x = 6

x = 3

Answer: The first term is 3. The common difference is 2.

8 0

3 years ago