All categories

3 years ago

The space inside a circle is the _ of the circle.

Mathematics

2 answers:

n200080 [17]3 years ago

8 0

Answer:

The space inside a circle is the area of the circle.

The border of the circle is the circumference of the circle.

oksian1 [2.3K]3 years ago

8 0

I think it is circumference

You might be interested in

Jack is trying to determine how he can save money by cutting his electricity bill. He is currently on a standard use plan for el

faust18 [17]

Jack's current monthly bill is
(1200 kWh +400 kWh)*($0.075/kWh) = $120

On the interval use plan, Jack's monthly bill would be
(1200 kWh)*($0.03/kWh) + (400 kWh)*($0.15/kWh) = $36 + 60 = $96

If Jack were to switch, his savings would be $120 -96 = $24.

The appropriate choice is ...
b. Jack should switch to the interval use plan, which would save him $24 per month.

7 0

3 years ago

Read 2 more answers

natali 33 [55]

Answer:

2.3 (or 2.293)

Step-by-step explanation:

25/10.99

6 0

3 years ago

Susie collected data on the favorite type of music of the students of two grades. The table below shows the relative frequencies

svp [43]

Answer:

In Grade 9, 20% of students liked classical music.

Step-by-step explanation:

Your table was hard to decipher, so I have recreated it here.

$\begin{array}{c|cccc}&\text{Pop}&\text{Classical}&\text{Heavy Metal}&\text{Total}\\\text{Grade 8}&0.50&0.10&0.40&1.00\\\text{Grade 9}&0.53&0.20&0.27&1.00\\\text{Total}&0.36&0.32&0.32&1.00\\\end{array}$

(a) In Grade 9, 53 students liked pop music.

FALSE. 53 % of the students liked pop music. We don't know how many students there were.

(b) In Grade 8, 60% of students liked heavy metal.

FALSE. 40 % of students liked heavy metal.

(c) In Grade 8, 10 students liked classical music.

FALSE. We don't know how many students there were.

(d) In Grade 9, 20% of students liked classical music.

TRUE. 0.20 means 20/100 or 20 %.

4 0

4 years ago

Starwatch camping site charge per person £5 please see attached

Travka [436]

Answer:

Dont understan what you are asking

Step-by-step explanation:

5 0

3 years ago

P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution

Vesna [10]

Sounds like you're asked to find $a$ such that

$\displaystyle\sum_{k=2}^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1$

In other words, find $a$ that satisfies

$a^2+a^3+a^4=1$

We can factorize this as

$a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0$

In order that $\mathbb P(k)$ describes a probability distribution, require that $\mathbb P(k)\ge0$ for all $k$ , which means we can ignore the possibility of $a=-1$ .

Let $a=y+\dfrac xy$ .

$a^3+a-1=\left(y+\dfrac xy\right)^3+\left(y+\dfrac xy\right)-1=0$
$\left(y^3+3xy+\dfrac{3x^2}y+\dfrac{x^3}{y^3}\right)+\left(y+\dfrac xy\right)-1=0$

Multiply both sides by $y^3$ .

$y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0$

We want to find $x\neq0$ that removes the quartic and quadratic terms from the equation, i.e.

$\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\dfrac13$

so the cubic above transforms to

$y^6-y^3-\dfrac1{27}=0$

Substitute $y^3=z$ and we get

$z^2-z-\dfrac1{27}=0\implies z=\dfrac{9+\sqrt{93}}{18}$
$\implies y=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}$
$\implies a=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}-\dfrac13\sqrt[3]{\dfrac{18}{9+\sqrt{93}}}$

6 0

4 years ago