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2 years ago

Sally was able to eat 5/8 of her lunch john are 75% of his lunch who are more of their meal

Mathematics

2 answers:

Nutka1998 [239]2 years ago

4 0

Answer:

john ate more because 5/8 is only 60 percent

garri49 [273]2 years ago

3 0

Answer:

John

Step-by-step explanation:

To be able to compare these easier, you'd turn the percentage 75% into a fraction. 75% equals 3/4. Then, you'd find a common denominator. Both 3/4 and 5/8 can be turned to have the denominator of 8. Since 5/8 already has the denominator of 8, you'd leave that alone. But, in order to get the denominator of 8 for 3/4, you'd have to multiply 4 times 2 to get 8. And, what you do to the bottom, you must do to the top, so 3 times 2 is 6. So 3/4 = 6/8.

SO, which is bigger Sally's 5/8, or John's 6/8? John ate more lunch.

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Un guardia forestal divisa, desde un punto de observación A, un incendio en la dirección N27° 10’ E. Otro guardia, que se encuen

Inga [223]

Answer:

la distancia del punto de observación A al incendio = 5.5 millas

la distancia del punto de observación B al incendio = 3.8 millas

Step-by-step explanation:

La expresión esquemática de la pregunta se puede ver en la imagen adjunta a continuación :

Del siguiente diagrama;

Usando la sine regla:

$\dfrac{sin \ F}{f} = \dfrac{sin \ A }{a}$

$\dfrac{sIn \ 79}{6} = \dfrac{sin \ 63}{a}$

a sin 79 = 6 sin 63

$a =\dfrac{6 \times sin \ 63}{sin \ 79}$

$a =\dfrac{6 \times 0.8910}{0.9816}$

a = 5.446 millas

la distancia del punto de observación A al incendio = 5.5 millas (a la décima más cercana)

$\dfrac{sin \ F}{f} = \dfrac{sin \ B}{b}$

$\dfrac{sin \ 79}{6} = \dfrac{sin \ 38}{b}$

b sin 79 = 6 sin 38

$b = \dfrac{6 \times sin \ 38}{sin \ 79}$

$b = \dfrac{6 \times0.6157 }{0.9816}$

b = 3.7635 millas

la distancia del punto de observación B al incendio = 3.8 millas (a la décima más cercana)

8 0

1 year ago

The number of hurricanes hitting the coast of Florida annually has a Poisson distribution with a mean of 0.8. Answer the followi

slava [35]

Answer:

a) $P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]$

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

b) $P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

Step-by-step explanation:

Let X the random variable that represent the number of hurricanes hitting the coast of Florida annualle. We know that $X \sim Poisson(\lambda=0.8)$

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 $\lambda=0.8$

$E(X)=\mu =\lambda=0.8$

Part a

For this case we want this probability: $P(X>2)$

And for this case we can use the complement rule like this:

$P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]$

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

Part b

Using the probability mass function we have:

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

3 0

2 years ago

vector w has magnitude 25.0 and direction angle 41.7 degrees. Calculate the horizontal and vertical components of w

11Alexandr11 [23.1K]

Consider the attached picture. It represents a vector, centered in the origin with angle 41.7 degrees, and a circumference with radius 25, centered in the origin.

As you can see, the horizontal and vertical components of the vector correspond to the lengths of segments OD and OC, respectively.

You might recognize this scenario from you trigonometry class: if the circle had radius 1, we would have

$OD=\cos(41.7),\quad OC=\sin(41.7)$