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serg [7]
2 years ago
8

Find the value of X. Round to the nearest tenth HELP ASAP!!!!

Mathematics
1 answer:
ZanzabumX [31]2 years ago
7 0

Answer: 22.5

Step-by-step explanation:

\sin 64^{\circ}=\frac{x}{25}\\\\x=25 \sin 64^{\circ}\\\\x \approx 22.5

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The answers on how I do these problems the correct way
dimulka [17.4K]

Step-by-step explanation:

so what you need to do is answer the questions with () and then do the other part

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One night Puvi spent 1/3 of his money on dinner and then 1/4 of the remaining money on dessert. He then had just enough left so
arsen [322]

Answer:

Step-by-step explanation:

Cost of two tickets = 2 * 30 = £60

Let the amount with Puvi = £x

Money spent on dinner = (1/3)*x

                                        = \dfrac{1}{3}x

Remaining \ money =x -  \dfrac{1}{3}x =\dfrac{3}{3}x-\dfrac{1}{3}x=\dfrac{2}{3}x

Money \ spent \ on \ dessert = \dfrac{1}{4} *\dfrac{2}{3}x =\dfrac{1}{6}x\\\\\\

Total money - money spent on dinner - money spent on dessert = 60

x -\dfrac{1}{3}x -\dfrac{1}{6}x=60\\\\\\\dfrac{6}{6}x-\dfrac{1*2}{3*2}-\dfrac{1}{6}x =60\\\\\\\dfrac{6}{6}x-\dfrac{2}{6}x-\dfrac{1}{6}x=60\\\\\\\dfrac{6-2-1}{12}x=60\\\\\\\dfrac{3}{12}x=60\\\\\\x=60*\dfrac{12}{3}=60*4=240

Money that Puvi had at the start of the night = £ 240

8 0
2 years ago
You invest $350 in an account with an interest rate of 1.2% compounded continuously. How much money would be in the account afte
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You will get about 1300 a year

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3 years ago
Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

P(X

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The cumulative distribution for this function is given by:

F(X) = 1- e^{-\lambda x}, x\ geq 0

We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

8 0
3 years ago
What is a reasonable measure for the angle shown?
dimaraw [331]
I would say 120 since it’s not at the end near 170

4 0
2 years ago
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