I am the distance around the circle- it's perimeter WHAT AM I?

A rectangular piece of plastic has an area of 10 square millimeters and a perimeter of 22 millimeters. What are the dimensions o

storchak [24]

The piece of plastic have dinmensions of 1mm by 10mm

5 0

3 years ago

Choose the correct correspondence RT

Marizza181 [45]

$RT$ corresponds to TR. correct option b.

Step-by-step explanation:

In the given parallelogram or rectangle , we have a diagonal RT . We need to find which side is in correspondence with side/Diagonal RT of parallelogram URST .

Side TU:

In triangle UTR , we see that TR is hypotenuse and is the longest side among UR & TU . So , TR can never be equal in length to UR & TU . So there's no correspondence of Side TU with RT.

Side TR:

Since, direction of sides are not mentioned here , we can say that TR & RT is parallel & equal to each other . So , TR is in correspondence with side/Diagonal RT of parallelogram URST .

Side UR:

In triangle UTR , we see that TR is hypotenuse and is the longest side among UR & TU . So , TR can never be equal in length to UR & TU . So there's no correspondence of Side UR with RT.

4 0

3 years ago

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300. The

zubka84 [21]

Answer:

the variance of the refund payment to the couple = 9463.394

Step-by-step explanation:

Given that :

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300.

It is possible that the couple won't be able to refund up 100 per day or more than 100 per day.

SO; let assume that the refund payment happens to be 0, 100,200, 300

Let X be the total refund payment on the seven day cruise.

We can say X = 0, if there is no rain on all 7 days.

$P(X = 0) = _nC_x * P^x * (1 - P)n-x$

$P(X = 0) = _7C_o * 0.2^0 * (1-0.2)^{7-0$

$P(X = 0) =1 * 1* (1-0.2)^{7$

$P(X = 0) =(0.8)^{7$

$P(X = 0) =0.2097152$

If it rains on any one day; then X = 100

$P(X = 100) = _nC_x * P^x * (1 - P)n-x$

$P(X = 100) = _7C_1 * 0.2^1 * (1-0.2)^{7-1$

$P(X = 0) =7 * 0.2* (1-0.2)^{6$

$P(X = 100) =7* 0.2* (0.8)^{6$

$P(X = 100) =0.3670016$

if it rains on any two day ; then X = 200

$P(X = 200) = _nC_x * P^x * (1 - P)n-x$

$P(X = 200) = _7C_2 * 0.2^2 * (1-0.2)^{7-2$

$P(X = 200) = 21 * 0.2^2 * (0.8)^{5$

$P(X = 200) = 0.2752512$

if it rains on any three day or more than that ; then X = 300

$P(X \ge 300) = 1 - P(X < 300) \\ \\ P(X \ge 300) = 1 - [P(X = 0) + P(X = 100) + P(X = 200)] \\ \\ P(X \ge 300) = 1 - [0.2097152 + 0.3670016 + 0.2752512] \\ \\ P(X \ge 300) = 0.148032$

Now; we have our probability distribution function as:

P(X = 0) = 0.2097152

P(X = 100) = 0.3670016

P(X = 200) = 0.2752512

P(X = 300) = 0.148032

In order to determine the variance of the refund payment to the couple; we use the formula:

variance of the refund payment to the couple $[Var X] =E [X^2] - (E [X])^2$

where;

$E[X^2] = \sum x^2 \times p \\ \\ E[X^2] = 0^2 * 0.2097152 + 100^2 * 0.3670016 + 200^2 * 0.2752512 + 300^2 * 0.148032 \\ \\ E[X^2] = 0 + 3670.016 + 11010.048+ 13322.88 \\ \\ E[X^2] =28002.944$

$(E [X]) = \sum x * p\\ \\ (E [X]) = 0 * 0.2097152 + 100 * 0.3670016 + 200 * 0.2752512 + 300 * 0.148032 \\ \\ (E [X]) = 0 + 36.70016 + 55.05024 + 44.4096\\ \\ (E [X]) = 136.16 \\ \\ (E [X])^2 = 136.16^2 \\ \\ (E [X])^2 = 18539.55$

NOW;

the variance of the refund payment to the couple = 28002.944 - 18539.55

the variance of the refund payment to the couple = 9463.394

7 0

3 years ago

A consistent system of equations is a system with __________.

mr_godi [17]

D. Intersecting lines and lines that have the same equation.

Hope this helps :)

4 0

3 years ago

Distribute 1/3(12y - 9)

allochka39001 [22]

3. 4y-3
because 1/3*12y - 1/3*9 = 4y - 3

7 0

3 years ago

I am the distance around the circle- it's perimeter WHAT AM I?​

I am the distance around the circle- it's perimeter WHAT AM I?