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2 years ago

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Mathematics

1 answer:

topjm [15]2 years ago

8 0

The Riemann sum with n = 6, taking the sample points to be midpoints is - 12.0625

<h3>What is Riemann sum?</h3>

Formula for midpoints is given as;

M = ∑0^n-1f((xk + xk + 1)/2) × Δx;

From the information given, we have the following parameters

x0 = 0
n = 6

xn = 3

Let' s find the parameters

Δx = (3 - 0)/6 = 0.5

xk = x0 + kΔx = 0.5k

xk+1 = x0 + (k +1)Δx

Substitute the values

= 0 + 0.5(k +1) = 0.5k - 0.5;(xk + xk+1)/2

We then have;

= (0.5k + 0.5k + 05.)/2

= 0.5k + 0.25.

Now f(x) = 2x^2 - 7

Let's find f((xk + xk+1)/2)

Substitute the value of (xk + xk+1)/2)

= f(0.5k+ 0.25)

= 2(0.5k + 0.25)2 - 7

Put values into formula for midpoint

M = ∑05[(0.5k + 0.25)2 - 7] × 0.5.

To evaluate this sum, use command SUM(SEQ) from List menu.

M = - 12.0625

A Riemann sum represents an approximation of a region's area from addition of the areas of multiple simplified slices of the region.

Thus, the Riemann sum with n = 6, taking the sample points to be midpoints is - 12.0625

Learn more about Riemann sum here:

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A distribution of values is normal with a mean of 188 and a standard deviation of 20.8. Find the probability that a randomly sel

padilas [110]

Answer:0.8413

Step-by-step explanation:

Mean= 188, Std Dev. =20.8, Z=(x-mean)/Std Dev

P(X less than 167.2)= P(Z less than (167.2-188)/20.8)

=P(Z less than (-20.8/20.8))

=P(z less than -1)= P(Z less 1)

The value of 1 in the normal distribution table is 0.3413

So we add 0.5 to 0.3413 =0.8413

6 0

3 years ago

1. If Jeremy found 80% of the collector cards and Tim found 41 of the 45 cards,

baherus [9]

Answer:

Tim has more

Step-by-step explanation:

41/45 when you type it into a calculator is 91% and 91%>80%

6 0

3 years ago

Please help me!<br> Find the number x such that f(x) =1

STatiana [176]

Answer:

Step-by-step explanation:

We have the piecewise function:

$f(x) = \left\{ \begin{array}{ll} -\frac{1}{2}x-1 & \quad x \leq -2 \\ x & \quad x > -2 \end{array} \right.$

And we want to find x such that f(x)=1.

So, let's substitute 1 for f(x):

$1 = \left\{ \begin{array}{ll} -\frac{1}{2}x-1 & \quad x \leq -2 \\ x & \quad x > -2 \end{array} \right.$

This has two equations. So, we can separate them into two separate cases. Namely:

$1=-\frac{1}{2}x-1\text{ or } 1=x$

Let's solve for x in each case.

Case I:

We have:

$1=-\frac{1}{2}x-1$

Add 1 to both sides:

$2=-\frac{1}{2}x$

Let's cancel out the fraction by multiplying both sides by -2. So:

$2(-2)=(-2)\frac{-1}{2}x$

The right side cancels:

$-4=x\\$

Flip:

$x=-4$

So, x is -4.

Case II:

We have:

$1=x$

Flip:

$x=1$

This is the solution for our second case.

So, we have:

$x_1=-4\text{ or } x_2=1$

Now, can check to see if we have to to remove solution(s) that don't work.

Note that x=-4 is the solution to our first equation.

The first equation is defined only if x is less than -2.

-4 <em>is </em>less than -2. So, x=-4 is indeed a solution.

x=1 is the solution to our second equation.

The second equation is defined only if x is greater than or equal to -2.

1 <em>is</em> greater than or equal to -2. So, x=1 is <em>also </em>a solution.

Therefore, our two solutions are:

$x_1=-4\text{ or } x_2=1$

Out of our answer choices, we can pick D.

And we're done!

7 0

3 years ago

If Upper X overbar equals 62, Upper S equals 8, and n equals 36, and assuming that the population is normally distributed, c

marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=62$ represent the sample mean

$\mu$ population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size

Part a: Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=36-1=35$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that $t_{\alpha/2}=2.72$

Now we have everything in order to replace into formula (1):

$62-2.72\frac{8}{\sqrt{36}}=58.373$

$62+2.72\frac{8}{\sqrt{36}}=65.627$

So on this case the 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

3 0

3 years ago

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia

ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

Xa = Xb

-6 + (3*t) = 7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

(6.8*t) = 13.4

t = 13.4/6.8

t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

4 0

3 years ago