for each store what is the ratio of number of can to the price 24 pk 2 for $9.00 12 pk 4 for $10.00 12 pk 3 for $9.00

What is the value of x in the figure? <br><br><br> x= ?

Otrada [13]

X = 68

Since the angle opposite of it is a right angle, we know that 22 + x has to equal 90 also.

So just take 90 - 22 = 68 degrees

6 0

3 years ago

Read 2 more answers

Need help with cross multiplying

saw5 [17]

So to solve for this, we need to set up proportional fractions, which I will help show you how to do.
First, if we are given an amount out of a total, we need to put it over x (if we are looking for the total). It looks like this:
12/x, 12 being the given number and x being the total.
If we are given the total but are looking for an amount, put the total at the bottom of the fraction (aka the denominator). It looks like this: x/16, 16 being the total amount and x being the amount out of the total.
We have a total of 40 test problems, so we can put our total at the bottom, x/40.
X is the amount of questions answered correctly (we are looking for x in the question).
We have answered 80% correct, so put 80% over 100 (100 being the total). It should look like this: 80/100.
Now we have our two fractions: x/40 & 80/100.
Set these up as an equation.
x/40 = 80/100.
Now this is where things may get tricky if you don't pay attention.
Multiply the numerator (the top number of a fraction) of x/40 by the denominator (the bottom number of a fraction) of 80/100.
Your product equation should look like this:
x times 100. This will give is 100x. Leave it at that.
Now, multiply the denominator of x/40 (the bottom number of the fraction) by the numerator (the top number of a fraction) of 80/100. It should look like this:
80 x 40. This will give us 3200.
Now set up our products as an equation.
100x = 3200.
To solve for x, divide both sides by 100.
3200/100 = 32.
x = 32.
I hope this helps and has taught you something!

4 0

3 years ago

A soda manufacturer knows that sales of soda increase during the summer. To make sure that they get a large portion of the sales

Maslowich

Answer:

a.) Probability = 1/5

b.) 10 soda

Step-by-step explanation:

Probability is a game of chance.

The initial chance = 1/12

If Lucky Lucy bought five sodas and won a free one with four of the caps. The probability of that happening will be 1/5

Having a greater than 50% chance of winning a free soda means having probability greater than half. That is,

Probability > 1/2

You would have to buy (12 - 2) soda. Which is equal to 10.

6 0

3 years ago

I need help ASAP please!! Giving brainliest!!!

Lemur [1.5K]

It’s the third option

4 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago