A. one solution <br> B. infinitely many solutions <br> C. no solution

The rectangle has length 8 cm and area 48 cm2.

marishachu [46]

Step-by-step explanation:

Given,

length of rectangle(l)= 8cm

area of rectangle(A) = 48cm2

breadth of rectangle(b) = ?

Perimeter of rectangle (P)=?

We know ,

Area of rectangle(A) = l×b

or, 48cm2 = 8cm×b

or, 48cm2 = 8bcm

or, 48cm2/8cm = b

or, 6cm = b

or, b = 6cm

therefore, b = 6cm

Perimeter of rectangle (P) = 2(l+b)

= 2(8cm+6cm)

= 2×14cm

= 28cm

therefore, Perimeter of rectangle(P) = 28cm

Now,

According to the question,

Perimeter of rectangle(P) = Perimeter of square(P)

So,

Perimeter of square(P) = 28cm

length of square(l) = ?

Area of square (A) = ?

We know,

Perimeter of square (P) = 4l

or, 28cm = 4l

or, 28cm/4 = l

or, 7cm = l

or, l = 7cm

therefore, l = 7cm

Now,

Area of square (A) = l^2

= (7cm)^2

= 7cm×7cm

= 49cm^2

therefore, area of square (A)= 49cm^2

3 0

3 years ago

Please help me i need help quick

RoseWind [281]

Answer:

Step-by-step explanation:

<u>Equating the equations</u> :

x - 4 = 4x + 2
3x = -6
x = -2

y = -2 - 4
y = -6

6 0

2 years ago

Read 2 more answers

X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y

andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

$\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}$

Part b.

Consider V(3X + 2Y)

$\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}$

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

$3 X+2 Y \sim N(18,77)$

Thus,

$P(3 X+2 Y < 18)=0.5$

Part d.

Consider P(3X + 2Y < 28)

$Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}$

$\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}$

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0

2 years ago

Identify the pattern and find the next number in the pattern. 0.8, 3.2 12.851.2

Savatey [412]

For this case we have the following sequence:
0.8
3.2
12.8
51.2
This sequence can be written as:
0.8 = 0.2 * 4 = 0.2 * 4 ^ 1
3.2 = 0.2 * 16 = 0.2 * 4 ^ 2
12.8 = 0.2 * 64 = 0.2 * 4 ^ 3
51.2 = 0.2 * 256 = 0.2 * 4 ^ 4
Therefore, we have that the generic expression for this case is:
an = 0.2 * 4 ^ n
Then, the next number is then:
a5 = 0.2 * 4 ^ 5 = 0.2 * 1024 = 204.8
answer
the next number in the pattern is 204.8

8 0

3 years ago

Need help ASAP!! I'll give brainliest to the correct answer!!!

Lapatulllka [165]

X = 12 inches so D is correct

3 0

3 years ago

A. one solution B. infinitely many solutions C. no solution