3.7 ÷ 1.1 = ? Round to the nearest hundredth. A. 0.297 B. 3.36 C. 4.07 D. 4.8

James folds a piece of paper in half several times,each time unfolding the paper to count how many equal parts he sees. After fo

Snezhnost [94]

Answer:

There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

The needed function is $y = 2 ^n$

Step-by-step explanation:

Let us assume the piece of paper James decides to fold is a SQUARE.

Now, let us assume:

n : the number of times the paper is folded.

y : The number of parts obtained after folds.

Now, if the paper if folded ONCE ⇒ n = 1

Also, when the pap er is folded once, the parts obtained are TWO equal parts.

⇒ for n = 1 , y = 2 ..... (1)

Similarly, if the paper if folded TWICE ⇒ n = 2

Also, when the paper is folded twice, the parts obtained are FOUR equal parts.

⇒ for n = 2 , y = 4 ..... (2)

⇒ $y = 2^2 = 2^n$

Continuing the same way, if the paper is folded SEVEN times ⇒ n = 7

So, $y = 2^ n = 2^7 = 128$

⇒ There are total 128 equal parts.

Lastly, if the paper is folded ELEVEN times ⇒ n = 11

So, $y = 2^ n = 2^{11} = 2048$

⇒ There are total 2048 equal parts.

Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.

And the needed function is $y = 2 ^n$

8 0

2 years ago

Two friends decide to go on a four hour ride on off road vehicles through mountainous terrain the graph represents their elevati

Aleksandr-060686 [28]

Answer:

Extrema: relative minimum (1.25,-3.25), relative maximums (3.25,10) and (0,0)

Zeros: (0,0), (2,0), and (4,0)

End behavior: As x approaches infinity, the function approaches negative infinity. As x approaches negative infinity, the function approaches negative infinity.

Intervals of increase and decrease: increasing on (-∞, 0) and (1.25, 3.25), decreasing on (0, 1.25) and (3.25, ∞)

Positive and negative intervals: positive on (2, 4), negative on (-∞, 0), (0, 2), and (4, ∞)

plato answer

Step-by-step explanation:

8 0

2 years ago

In △ABC, m∠ABC=40°,

Alchen [17]

Let m∠CLN = x. Then m∠ALM = 3x, and m∠A = 90°-x, m∠C = 90°-3x.

The sum of angles of ∆ABC is 180°, so we have

... 180° = 40° + m∠A + m∠C

Using the above expressions for m∠A and m∠C, we can write ...

... 180° = 40° + (90° -x) + (90° -3x)

... 4x = 40° . . . . . . . . . add 4x-180°

... x = 10°

From which we conclude ...

... m∠C = 90°-3x = 90° - 3·10° = 60°

The ratio of CN to CL is

... CN/CL = cos(∠C) = cos(60°)

... CN/CL = 1/2

so ...

... CN = (1/2)CL

5 0

2 years ago

W = 3x + 7y solve for y

mario62 [17]

Answer:

The value of the equation $y=\frac{W-3x}{7}$ .

Step-by-step explanation:

Consider the provided equation.

$W = 3x + 7y$

We need to solve the provided equation for y.

Subtract 3x from both side.

$W-3x= 3x-3x+ 7y$

$W-3x=7y$

Divide both sides by 7.

$\frac{7y}{7}=\frac{W-3x}{7}$

$y=\frac{W-3x}{7}$

Hence, the value of the equation is $y=\frac{W-3x}{7}$ .

8 0

2 years ago

After calculating the sample size needed to estimate a population proportion to within 0.05, you have been told that the maximum

Svetlanka [38]

Answer: 40000

Step-by-step explanation:

The formula to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$ , where p is the prior estimate of the population proportion.

Here we can see that the sample size is inversely proportion withe square of margin of error.

i.e. $n\ \alpha\ \dfrac{1}{E^2}$

By the equation inverse variation, we have

$n_1E_1^2=n_2E_2^2$

Given : $E_1=0.05$ $n_1=1000$

$E_2=0.025$

Then, we have

$(1000)(0.05)^2=n_2(0.025)^2\\\\\Rightarrow\ 2.5=0.000625n_2\\\\\Rightarrow\ n_2=\dfrac{2.5}{0.000625}=4000$

Hence, the sample size will now have to be 4000.

6 0

2 years ago