Ask question

Ask question

All categories

3 years ago

5

is 0/8 undefined? I know 8/0 is undefined, but I need to know if 0/8 is undefined as well. (Please answer like so: 0/8 is/is not

undefined)

2 answers:

igomit [66]3 years ago

6 0

Answer:

0/8 is not undefined.

Step-by-step explanation:

Since, zero can be divided by any number except infinity because it is an indeterminate value.

Also, When we divide zero by a number ( no matter how big or small that number is ), the quotient is always zero,

And, zero is a determined value ( or defined value ),

Thus, 0/8 = 0,

⇒ 0/8 is defined.

Note: The division by zero is undefined.

Nata [24]3 years ago

3 0

0/8 is 0
8/0 is undefined :)

You might be interested in

Find the product <br> -1/2y(2y^3-8)

blondinia [14]

Answer:

-y^4+4y

Step-by-step explanation:

3 0

3 years ago

Create a matrix that is equal to F+G. The first matrix below is named F and the second matrix below is named G. Name the new mat

likoan [24]

F+G:

$F+G=\begin{bmatrix}{-1.8} & {-8.6} & {} \\ {2.85} & {-1.4} & {} \\ {-1.8} & {5.1} & {}\end{bmatrix}+\begin{bmatrix}{1.32} & {-1.9} & {} \\ {2.25} & {0.0} & {} \\ {-6.2} & {1.4} & {}\end{bmatrix}$

Then, add the elements that occupy the same position:

$H=\begin{bmatrix}{-1.8+1.32} & {-8.6+(-1.9)} & {} \\ {2.85+2.25} & {-1.4+0.0} & {} \\ {-1.8+(-6.2)} & {5.1+1.4} & {}\end{bmatrix}$

Solve

$H=\begin{bmatrix}{-0.48} & {-10.5} & {} \\ {5.1} & {-1.4} & {} \\ {-8} & {6.5} & {}\end{bmatrix}$

So, we find the element at address h31:

$H=\begin{bmatrix}{h11} & {h12} & {} \\ {h21} & {h22} & {} \\ {h31} & {h32} & {}\end{bmatrix}$

In this case, position h31 is - 8.0

8 0

1 year ago

To solve a problem, we often _________ the given information into algebraic expressions and equations.

igomit [66]

I think the answer is we often insert the given information.

7 0

3 years ago

Read 2 more answers

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago

The number 1,000,000 is a exponent of 10

kaheart [24]

1,000,000 = 10^6 because we have six 0;

3 0

3 years ago

Read 2 more answers