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3 years ago

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How to solve this math problem

1 answer:

Bumek [7]3 years ago

3 0

Answer:

5.833

Step-by-step explanation:

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3 years ago

3 geometry questions, 30 points!

polet [3.4K]

Answer:

A ) The value of x for the given circle with chords and center is 8.3

B) The circumference of circle with chords 1.2 cm and 0.5 cm is 4.082

Step-by-step explanation:

Given two figures :

<u>For Figure first </u>

A circle with center y , having two chords FM and NM

FM = 5 x

MN = 2 x + 25

Now from theorem of circle ,

Chords equidistant from center of circle are equal in length

I.e distance of chord MN from center y and distance of FM from center y are equal

So, FM = MN

Or, 5 x = 2 x + 25

Or, 5 x - 2 x = 25

Or, 3 x = 25

∴ x = $\frac{25}{3}$ = 8.33

<u>For figure second</u>

The length of two adjacent chords of circle is 1.2 cm and 0.5 cm

Let the center of circle = O

Length of chord AB = 1.2 cm

Length of chord BC = 0.5 cm

As both chords are at 90° to each other

So The Length of diameter of circle AC = $\sqrt{AB^{2}+BC^{2}}$

Or, The Length of diameter of circle AC = $\sqrt{1.2^{2}+0.5^{2}}$

Or, The Length of diameter of circle AC = $\sqrt{1.44+0.25}}$

Or, The Length of diameter of circle AC = $\sqrt{1.69}$

∴ The Length of diameter of circle AC = 1.3 cm

So, Circumference of circle = $\pi d$

Or, Circumference of circle = 3.14 × 1.3

∴ Circumference of circle = 4.082 cm

Hence,

A ) The value of x for the given circle with chords and center is 8.3

B) The circumference of circle with chords 1.2 cm and 0.5 cm is 4.082 Answer

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3 years ago

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Which of the following transforms y = x' to the graph of y = (x + 5)22

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2 years ago

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

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

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

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

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

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

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