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4 years ago

11

What is the distance between –125 and 78?

1 answer:

nexus9112 [7]4 years ago

8 0

The disstance between -125 and 78 is 47

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If triangle DEF is an isosceles triangle with DE is congruent to EF, find x and the measure of each side

Rudik [331]

Step-by-step explanation:

Given that,

DE = 8x - 13

EF = 5x + 17

DF = x + 21

Also,

DE = EF

which means that,

8x - 13 = 5x + 17

8x - 5x = 17 + 13

3x = 30

x = 30/3

x = 10

Now,

DE = 8x - 13 = 8×10 - 13 = 80 - 13 = 67cm

EF = 5x + 17 = 5×10 + 17 = 50 + 17 = 67cm

DF = x + 21 = 10 + 21 = 31cm

3 0

3 years ago

The cost of renting roller blades is $4 plus $3.50 for each hour that the roller blades are rented. Write an expression to repre

Pavel [41]

Answer:

The expression is equal to $3.50h+4$

Step-by-step explanation:

Let

x ----> the number of hours that the roller blades are rented

y ----> the cost of renting roller blades in dollars

we know that

The linear equation in slope intercept form is equal to

$y=mx+b$

where

m is the slope or unit rate of the linear equation

b is the y-intercept or initial value of the linear equation

In this problem we have

The slope is equal to

$m=\$3.50\ per\ hour$

The y-intercept is equal to

$b=\$4$ ----> value of y when the value of x is equal to zero

substitute

$y=3.50x+4$

For x=h hours

substitute

$y=3.50h+4$

3 0

3 years ago

If the image above is rotated about the y-axis, which of the following images best represents the result?

swat32

Answer:

the answer would be x because it has a larger width than y

6 0

3 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

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

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

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

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

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

4 0

3 years ago

Please help with this, Thank You.

Dmitry [639]

Answer:

$x = 12$

Step-by-step explanation:

$3x + 9 = 5x - 15$

$2x = 24$

$x = 12$

4 0

2 years ago