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Akimi4 [234]
3 years ago
6

I would really appreciate some help... branliest to whoever can

Mathematics
1 answer:
Margaret [11]3 years ago
5 0

Answer:

1. 6 to the power of 8

2. 9765625

3. 2 to the power of 4

4. 1296

5. I do not know 5 sorry.

6. 117649

Step-by-step explanation:

You might be interested in
A company installs 5,000 light bulbs. the lifetimes of the lightbulbs are approximately normally distributed with a mean of 500
Sergio039 [100]
First, determine the z-score of 675. 
                               z = (675 - 500) / 100 = 1.75
The z-score of 500 is,
                               z = 0. 
Subtracting the z-scores will give us 1.75. This is equal to 0.9599. 
                        = 0.9599 - 0.5 = 0.4599
Multiplying this to the given number of light bulbs,
                             n = 0.4599 x 5000 = 2299.5
Therefore, there is approximately 2300 light bulbs expected to last between 500 to 675 hours.
5 0
2 years ago
A disco ball is shaped like a sphere with a diameter of 16 inches. To the nearest square inch, what is the surface of the disco
Mademuasel [1]
If the diameter is 16 inches, then its radius is half that, r = 8.

\bf \textit{surface area of a sphere}\\\\
A=4\pi r^2\quad 
\begin{cases}
r=radius\\
-----\\
r=8
\end{cases}\implies A=4\pi 8^2
5 0
3 years ago
Solve 9≥x+15.<br><br> The solution is ___
Lana71 [14]

Answer: 6

Step-by-step explanation:

15 - 9 = 6 to the quadratics x inequalities formula x b to the power of \int\limits^a_b {x} \, dx

4 0
3 years ago
Read 2 more answers
Plssss help<br><br>use the property that it says to solve the proportion ty​
Kay [80]

Answer:

the answer is 45 over 6

Step-by-step explanation:

A shortcut is "2 x _____ is 6", which is 3.

Multiply 15 by 3 also, to get 45,

45/6

7 0
2 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
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