What are the factors of 5

WHEN ADDING THE NUMBERS 5.7 + 2.31 EXPLAIN HOW YOU WOULD LINE UP THE NUMBERS VERTICALLY ONE ON TOP OF THE OTHER.IN ORDER TO SLOV

Volgvan

Vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

4 0

3 years ago

FIRST ANSWER GETS BRAINLIEST

Genrish500 [490]

Answer:

c

Step-by-step explanation:

im DORA

7 0

2 years ago

X^2 + y^2 = 8 X-y=0 Select all of the following that are solutions to the system shown

ludmilkaskok [199]

X^2 + y^2 = 8
X-y=0 so x = y

replace x = y into X^2 + y^2 = 8

y^2 + y^2 = 8

2y^2 = 8

y^2 = 8/2

y^2 = 4

y = - 2 and y = 2

because x = y

so x = - 2 and x = 2

solutions:

x= - 2 and x = + 2

y= - 2 and y = + 2

5 0

2 years ago

Can someone give me an example on a Riemann Sum and like how to work through it ? I want to learn but I don’t understand it when

Georgia [21]

Explanation:

A Riemann Sum is the sum of areas under a curve. It approximates an integral. There are various ways the area under a curve can be approximated, and the different ways give rise to different descriptions of the sum.

A Riemann Sum is often specified in terms of the overall interval of "integration," the number of divisions of that interval to use, and the method of combining function values.

Example Problem

For the example attached, we are finding the area under the sine curve on the interval [1, 4] using 6 subintervals. We are using a rectangle whose height matches the function at the left side of the rectangle. We say this is a left sum.

When rectangles are used, other choices often seen are right sum, or midpoint sum (where the midpoint of the rectangle matches the function value at that point).

Each term of the sum is the area of the rectangle. That is the product of the rectangle's height and its width. We have chosen the width of the rectangle (the "subinterval") to be 1/6 of the width of the interval [1, 4], so each rectangle is (4-1)/6 = 1/2 unit wide.

The height of each rectangle is the function value at its left edge. In the example, we have defined the function x₁(j) to give us the x-value at the left edge of subinterval j. Then the height of the rectangle is f(x₁(j)).

We have factored the rectangle width out of the sum, so our sum is simply the heights of the left edges of the 6 subintervals. Multiplying that sum by the subinterval width gives our left sum r₁. (It is not a very good approximation of the integral.)

The second and third attachments show a right sum (r₂) and a midpoint sum (r₃). The latter is the best of these approximations.

_____

Other Rules

Described above and shown in the graphics are the use of rectangles for elements of the summation. Another choice is the use of trapezoids. For this, the corners of the trapezoid match the function value on both the left and right edges of the subinterval.

Suppose the n subinterval boundaries are at x0, x1, x2, ..., xn, so that the function values at those boundaries are f(x0), f(x1), f(x2), ..., f(xn). Using trapezoids, the area of the first trapezoid would be ...

a1 = (f(x0) +f(x1))/2·∆x . . . . where ∆x is the subinterval width

a2 = (f(x1) +f(x2))/2·∆x

We can see that in computing these two terms, we have evaluated f(x1) twice. We also see that f(x1)/2 contributes twice to the overall sum.

If we collapse the sum a1+a2+...+an, we find it is ...

∆x·(f(x0)/2 + f(x1) +f(x2) + ... +f(x_n-1) + f(xn)/2)

That is, each function value except the first and last contributes fully to the sum. When we compute the sum this way, we say we are using the trapezoidal rule.

If the function values are used to create an approximating parabola, a different formula emerges. That formula is called Simpson's rule. That rule has different weights for alternate function values and for the end values. The formulas are readily available elsewhere, and are beyond the scope of this answer.

_____

Comment on mechanics

As you can tell from the attachments, it is convenient to let a graphing calculator or spreadsheet compute the sum. If you need to see the interval boundaries and the function values, a spreadsheet may be preferred.

8 0

2 years ago

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

iris [78.8K]

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is $\\ P(z>0.48) = P(x>533.2) = 0.3156$

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a sample mean. The distribution for sample means follows a normal distribution with mean $\\ \mu$ and standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ . Mathematically

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

For values of the sample $\\ n \ge 30$ , no matter the distribution the data come from.

And the variable z follows a standard normal distribution, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

That is

$\\ z \sim N(0, 1)$

We have a variance of 3364. That is, a standard deviation of

$\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58$

The population mean is

$\\ \mu = 530$

The sample size is $\\ n = 75$

The sample mean is $\\ \overline{x} = 533.2$

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}$

$\\ z = \frac{3.2}{\frac{58}{8.66025}}$

$\\ z = \frac{3.2}{6.69726}$

$\\ z = 0.47780$

With this value of z we can consult a cumulative standard normal table (or use some statistic program) to find the cumulative probability for z (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

$\\ P(z$

However, in the question we are asked for $\\ P(z>0.48) = P(x>533.2)$ . As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

$\\ P(z>0.48) = 1 - P(z$

Thus

$\\ P(z>0.48) = 1 - 0.68439$

$\\ P(z>0.48) = 0.31561$

Rounding to four decimal places, we have

$\\ P(z>0.48) = 0.3156$

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) $\\ P(z>0.48) = P(x>533.2) = 0.3156$ .

5 0

3 years ago