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3 years ago

Round 0.868 to the nearest tenth.

Mathematics

2 answers:

Anna11 [10]3 years ago

4 0

Answer: 0.9

Step-by-step explanation:

Nitella [24]3 years ago

3 0

Answer: 0.9

If we round 0.868 then the answer would be 0.9

When we round and the number is 5, 6, 7, 8, 9, then you will go ahead and round it. If the number is below 5 then do not round the number.

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Researchers gave 40 index cards to a waitress at an Italian restaurant in New Jersey. Before delivering the bill to each custome

rjkz [21]

The test statistic and p-value of the given data are 6.274 and 0.0001 respectively.

<h3>Test Statistic</h3>

The test statistic can be calculated using the formula below

$t = \frac{x-\mu}{\sigma / \sqrt{n} }$

Solving for the mean and standard deviation, we can substitute the values into the above equation which will be

$t = \frac{x-\mu}{\sigma / \sqrt{n} } = 6.274$

<h3>P-Value</h3>

Using the data from the test statistic, we can calculate the p-value of the data

$p-value = p(z < 6.274)= 0.0001 = 0.00$

From the calculation above, the test statistic and p-value of the given data are

6.274
0.0001

Learn more on test statistic and p-value here;

brainly.com/question/4621112

3 0

2 years ago

What is the solution of 4+ V5x+88 – X+

N76 [4]

Answer: no solution

Step-by-step explanation:

4 0

3 years ago

Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

$y'' - 3y' + 2y = e^x$

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

6 0

1 year ago

PLEASE ANSWER ASAP WILL MARK BRAINLIEST

Olin [163]

Answer:

First choice

Step-by-step explanation:

4 0

2 years ago

Which expression is equivalent to 2w?

kotykmax [81]

Answer:

2w - w would be equal to w. w + 2 is just w + 2

Step-by-step explanation:

please give brainlist if this helped

8 0

3 years ago