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3 years ago

9

Use the distributive property to remove the parentheses.

1 answer:

nlexa [21]3 years ago

8 0

Answer:

Step-by-step $12c^{3} -40c^{8}$ explanation:

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LMNO~QRST find the value of x.

adelina 88 [10]

Since both trapezoids are similar, so LMNO is a scaled version of QRST by some magnitude
so we can easily use the concept of ratios to find x
in other words: x÷2=6÷5
so x=6×2÷5=12/5=2.4
x=2.4

4 0

3 years ago

Jamie says that if she knows only the measure of one angle in an isosceles triangle, she can always determine the measures of th

lozanna [386]

Answer:

Jamie is correct

Step-by-step explanation:

Jamie is correct.

Example: isosceles triangle ABC AB=AC

∠B = ∠C ∠A + ∠B + ∠C = 180°

if ∠A = x ∠B = ∠C = 1/2 * ( 180° - x)

if ∠B or ∠C = x ∠A = 180° - 2x

6 0

4 years ago

Read 2 more answers

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

b) $P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452$

c) $P(X \geq 8) = 1-P(X$

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

$P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465$

$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

$P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298$

$P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452$

c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

$P(X \geq 8) = 1-P(X$

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago

On a city map, a rectangular park has a length of 2 in. If the actual length and width of the park are 400 ft and 100 ft, res

pshichka [43]

Answer:

0.5 inches

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What would the area of an actual pool be in the scale from the drawing is 3:1000

elena55 [62]

Answer:

If this is the complete question them the answer is 1250 m²

Step-by-step explanation:

the pool in the drawing is 15x7.5 and the scale is 3/1000 so if the size in the drawing is 3 then the actual is 1000.

x=15*1000/3=5000cm which would be 50m

x=7.5*1000/3=2500cm which is 25m

the actual area of the pool is 50*25=1250 m²

8 0

3 years ago