Is the function negative over (-6,-2)

The following cube has a volume of 8000 cm3. Find the length of one side of the cube.

Artemon [7]

Vcube = Ledge^3
8000cm^3 = Ledge^3

use the cube root of each side and you have the measure of an edge in cm.

8000^(1/3) = 20
every edge is 20 cm

20^2 = 400
The area of each face is 400 cm^2.

6 0

3 years ago

Please help! This is due soon :(

Papessa [141]

Answer:

what grade us it maybe u can help you but if I cant I'm sorru

8 0

3 years ago

Use Cramers Rule to solve the system 2x - 3y = 11 -6x + 8y = 34

Nastasia [14]

Step-by-step explanation:

given :

2x - 3y = 11

-6x + 8y = 34

find : the solutions of the system by using Cramers Rule.

solutions:

in the matrix 2x2 form =>

[ 2 -3] [x] [11]

=

[-6 8] [ y] [34]

D =

| 2 -3 |

|-6 8 |

= 8×2 - (-3) (-6)

= 16-18 = -2

Dx = | 11 -3 |

| 34 8 |

= 11×8 - (-3) (34)

= 88 + 102

= 190

Dy = | 2 11 |

|-6 34 |

= 2×34 - (-6) (11)

= 68 + 66

= 134

x = Dx/D = 190/-2 = -95

y = Dy/D = 134/-2 = -67

the solutions = {-95, -67}

8 0

3 years ago

Read 2 more answers

Find the annual percentage rate of change for the population of Oregon. In 2000, there were 4.8

MrRa [10]

Answer:

The annual percentage rate of change for the population of Oregon is of 8.125%.

Step-by-step explanation:

Total percentage change:

Change multiplied by 100 and divided by the initial value.

Change: 8.7 - 4.8 = 3.9 million

Initial value: 4.8

Percentage change: 3.9*100/4.8 = 81.25%

Annual percentage rate of change

81.25% during 10 years(from 2000 to 2010).

So, per year

81.25%/10 = 8.125%

The annual percentage rate of change for the population of Oregon is of 8.125%.

5 0

3 years ago

A professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their class

KIM [24]

Answer:

We conclude that seniors skip more than 2% of their classes at 0.01 level of significance.

Step-by-step explanation:

We are given that a professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their classes.

He randomly samples a group of seniors and out of 2521 classes, the group skipped 77.

Let p = percentage of seniors who skip their classes.

So, Null Hypothesis, $H_0$ : p $\leq$ 2% {means that seniors skip less than or equal to 2% of their classes}

Alternate Hypothesis, $H_A$ : p > 2% {means that seniors skip more than 2% of their classes}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of seniors who skipped their classes = $\frac{77}{2521}$

n = sample of classes = 2521

So, test statistics = $\frac{\frac{77}{2521} -0.02}{{\sqrt{\frac{\frac{77}{2521}(1-\frac{77}{2521})}{2521} } } } }$

= 3.08

The value of the test statistics is 3.08.

Now at 0.01 significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical value of z as 2.3263 < 3.08, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that seniors skip more than 2% of their classes.

6 0

4 years ago