Suppose the length and the width of the rectangle are doubled. What

So how do i so how do I solve to get this answer

Elina [12.6K]

Step-by-step explanation:

First, You would solve what you can which would be the equation on the left Which will get you 4096. After that its just solving for P

4096=(4^p)^2

64=4^p (you would take the square root of 4096 to get 64)

Then, Since 64 is a perfect Cube root of 4, P will equal 3

(Please award this Brainliest and Thank the skeleton in your closet)

8 0

3 years ago

Drag each tile to the correct box

ValentinkaMS [17]

Answer:

Arranging the solutions from the least to the highest;

1. 1.743 x 10⁻²

2. 3.626 x 10⁻²

3. 4.64 x 10⁻²

4. 3.162 x 10²

5. 4.214 x 10³

Step-by-step explanation:

The solutions of the mathematical expression are calculated as follows;

$1. \ \ (4.3\times 10^6)(9.8\times 10^{-4}) = (4.3\times 9.8\times 10^{6-4}) = (42.14\times 10^2)= 4.214 \times 10^3$

$2. \ \ (2.9\times 10^7)(1.6\times 10^{-9}) = (2.9\times 1.6\times 10^{7-9}) = 4.64\times 10^{-2}$

$3. \ \ \frac{(4.7 \times 10^3)(8.6\times 10^{-7})}{(3.8\times 10^{-4})(6.1 \times 10^2)} = \frac{(4.7\times 8.6\times 10^{3-7})}{(3.8\times 6.1\times 10^{-4+2})} = \frac{4.042 \times 10^{-3}}{2.318\times 10^{-1}} = 1.743\times 10^{-2}$

$4. \ \ (4.9\times 10^3)(7.4\times 10^{-6}) = (4.9\times 7.4\times 10^{3-6}) = 3.626\times 10^{-2}$

$5. \ \ \frac{(3.9 \times 10^5)(8.7\times 10^{-3})}{(3.7\times 10^{-5})(2.9 \times 10^8)} = \frac{(3.9\times 8.7\times 10^{5-3})}{(3.7\times 2.9\times 10^{-4 +8 })} = \frac{3.393 \times 10^{-1}}{1.073\times 10^{-3}} = 3.162\times 10^{2}$

Arranging the solutions from the least to the highest;

1. 1.743 x 10⁻²

2. 3.626 x 10⁻²

3. 4.64 x 10⁻²

4. 3.162 x 10²

5. 4.214 x 10³

4 0

3 years ago

Read 2 more answers

I've been stuck on this for the past hour. Can someone tell me how to solve it?

arlik [135]

Answer:

The answer is 2.5

Step-by-step explanation:

Your welcome! I sometimes get stuck on problems like you too for hours until I get it.

6 0

3 years ago

Read 2 more answers

How is a trapezoid different from the other four types of quadrilaterals you have learned about?

navik [9.2K]

It could be because of its shape? but im not all the way sure (plz mark me as brainliest and hit the thank u button) ps im sorry if I was wrong

5 0

3 years ago

Read 2 more answers

The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with mean 1252

stich3 [128]

Answer:

a) P ( 1100 < X < 1400 ) = 0.755

b) P ( X < 1000 ) = 0.755

c) proportion ( X > 1200 ) = 65.66%

d) 5.87% percentile

Step-by-step explanation:

Solution:-

- Denote a random variable X: The number of chocolate chip in an 18-ounce bag of chocolate chip cookies.

- The RV is normally distributed with the parameters mean ( u ) and standard deviation ( s ) given:

u = 1252

s = 129

- The RV ( X ) follows normal distribution:

X ~ Norm ( 1252 , 129^2 )

a) what is the probability that a randomly selected bag contains between 1100 and 1400 chocolate chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

P ( x1 < X < x2 ) = P ( [ x1 - u ] / s < Z < [ x2 - u ] / s )

- Taking the limits x1 = 1100 and x2 = 1400. The standard normal values are:

P ( 1100 < X < 1400 ) = P ( [ 1100 - 1252 ] / 129 < Z < [ 1400 - 1252 ] / 129 )

= P ( - 1.1783 < Z < 1.14728 )

- Use the standard normal tables to determine the required probability defined by the standard values:

P ( -1.1783 < Z < 1.14728 ) = 0.755

Hence,

P ( 1100 < X < 1400 ) = 0.755 ... Answer

b) what is the probability that a randomly selected bag contains fewer than 1000 chocolate chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

P ( X < x2 ) = P ( Z < [ x2 - u ] / s )

- Taking the limit x2 = 1000. The standard normal values are:

P ( X < 1000 ) = P ( Z < [ 1000 - 1252 ] / 129 )

= P ( Z < -1.9535 )

- Use the standard normal tables to determine the required probability defined by the standard values:

P ( Z < -1.9535 ) = 0.0254

Hence,

P ( X < 1000 ) = 0.755 ... Answer

(c) what proportion of bags contains more than 1200 chocolate chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

P ( X > x1 ) = P ( Z > [ x1 - u ] / s )

- Taking the limit x1 = 1200. The standard normal values are:

P ( X > 1200 ) = P ( Z > [ 1200 - 1252 ] / 129 )

= P ( Z > 0.4031 )

- Use the standard normal tables to determine the required probability defined by the standard values:

P ( Z > 0.4031 ) = 0.6566

Hence,

proportion of X > 1200 = P ( X > 1200 )*100 = 65.66% ... Answer

d) what is the percentile rank of a bag that contains 1050 chocolate chips?

- The percentile rank is defined by the proportion of chocolate less than the desired value.

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

P ( X < x2 ) = P ( Z < [ x2 - u ] / s )

- Taking the limit x2 = 1050. The standard normal values are:

P ( X < 1050 ) = P ( Z < [ 1050 - 1252 ] / 129 )

= P ( Z < 1.5659 )

- Use the standard normal tables to determine the required probability defined by the standard values:

P ( Z < 1.5659 ) = 0.0587

Hence,

Rank = proportion of X < 1050 = P ( X < 1050 )*100

= 0.0587*100 %

= 5.87 % ... Answer

6 0

3 years ago