Can someone pls help me ? I really need help

6 squared + 8squared = 100 what would x^2 (squared) divides by 100 be ?

Keith_Richards [23]

If you are trying to say what is square root of 100 then it is 10. Like 10^2 equals 100.

4 0

3 years ago

Give a big-o estimate for the number of operations (where an operation is an addition or a multiplication) used in this segment

Marianna [84]

We have a "rectangular" double loop, meaning that both loops go to completion.
So there are 3*4=12 executions of t:=t+ij.

Assuming two operatiions per execution of the innermost loop, (i.e. ignoring the implied additions in increment of subscripts), we have 12*2=24 operations in all.

Here the number of operations (+ or *) is exactly known (=24).

Big-O estimates are used for cases with a varying scale of operations, governed by a variable (usually n) to indicate the sensitivity of the number of operations relative to a change in the size of n.

Here we do not have a scale, nor n is defined. The number of operations is constant and known at 24. So a variable is required to find the big-O estimate.

6 0

4 years ago

Let x denote the lifetime of a mcchine component with an exponential distribution. The mean time for the component failure is 25

aliina [53]

Answer:

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The mean time for the component failure is 2500 hours.

This means that $m = \frac{2500}, \mu = \frac{1}{2500} = 0.0004$

What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?

The standard deviation of the exponential distribution is the same as the mean, so this is P(X > 5000).

$P(X > x) = e^{-0.0004*5000} = 0.1353$

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

4 0

3 years ago

I need to know how to solve -4-(2+4(-6)-4-(2^2-5*2))

sweet-ann [11.9K]

Answer:

16

Step-by-step explanation:

This problem requires PEMDAS

Parentheses ( )

Exponents ^

Multiplication

Division

Add

Subtract

Start by solving anything in parentheses. There's an exponent within the parentheses, so we change that 2^2 into 4 and also make sure to multiply 5 x 2 before subtracting.

-4 - (2 + -24 - 4 - (4-10))

-4 - (2 + -24 - 4 - (-6))

Again, solve parentheses first.

-4 - (-22 - 4 - (-6))

-4 - (-26 + 6)

-4 - (-20)

-4 + 20

Answer is 16

4 0

3 years ago

Multiply (5 + 3i)(5 − 3i) 16 + 15i 16 − 15i 16 34

Annette [7]