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3 years ago

15

A survey of 1950 people found that 39% preferred chocolate ice cream to vanilla. About how many people preferred chocolate ice c

ream according to the survey?

2 answers:

Diano4ka-milaya [45]3 years ago

8 0

Answer:

its c

Step-by-step explanation:

FrozenT [24]3 years ago

6 0

N(U) = 1950
39% of them preferred chocolate ice-cream.
Then,
no. Of people who preferred chocolate ice-cream= 39% of 1950
=39/100 *1950
= 760.5

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The formula for the circumference of a circle is c= 2pi r, where r is the radius. Rearrange the formula to solve for our and sel

ratelena [41]

Answer: r = C/(2pi)

This is the same as writing $r = \frac{C}{2\pi}$

=========================================================

Explanation:

pi is some number (approximately 3.14) which means 2pi is also some number (roughly 6.28)

Saying c = 2pi*r means we have 2pi times r, and the result is the circumference c. To isolate r, we'll undo the multiplication. We'll undo it by dividing both sides by 2pi like so

C = 2pi*r

C/(2pi) = r

r = C/(2pi)

in which we can write it like $r = \frac{C}{2\pi}$

So whatever C is, we divide it over 2pi (aka roughly 6.28) to get the radius.

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Extra Info (Optional Section):

As an example, let's say the circumference of the circle is 628 feet. This means the distance around the circle is 628 feet. So C = 628 would lead to...

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2 years ago

(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho

Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

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3 years ago

Which angles are adjacent to each other? Question is located in the attached file (image)

daser333 [38]

Answer:

∠ 1 and ∠2

Step-by-step explanation:

Adjacent angles share a common ray or side between them, and have the same vertex.

Therefore, the adjacent angles from the given image are: ∠ 1 and ∠2.

The other given options are not adjacent angles because they do not have the same common vertex and ray.

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2 years ago