All categories

3 years ago

Find the area of the shape below. * PLSS HELP

Mathematics

2 answers:

Travka [436]3 years ago

6 0

Answer:

because the formula for the triangle is 1/2×b×h

and the formula for the rectangle nos l×w and then add those.

ki77a [65]3 years ago

4 0

Answer:

A = 60yd

Step-by-step explanation:

First "separate" the rectangle and the triangle in your head.

Now calculate the area of the rectangle and then the area of the triangle.

Rectangle: 6*8 = 48

Triangle: (6*4)/2 = 24/2 = 12

Now we add them together:

48 + 12 = 60yd

You might be interested in

What lengths are proportional to 4.5 and 1.5

alexgriva [62]

3

1.5 and 4.5 all are divisible by 1.5. Then you just have to find another number that is as well. I chose 3. Theres also 6, 7.5,9,10.5,12

8 0

3 years ago

Explain why the X coordinates of the points where the graphs of the equations y=2^-x and y=4^x+3 intersects

sveta [45]

Hello,

y=2^(-x)
y=2^(2x)+3

==>2^(2x)+3=1/2^x
==>2^(3x)+3*2^x-1=0 (1)
Let's assume u=2^x
(1)==>u^3+3*u-1=0

which as 3 roots
u=0.322185354626 or
u = -0.161092677313 + i1.754380959784 or
u = -0.161092677313 - i1.754380959784.

Let's take the real solution

0.322185354626=2^x
==>x=ln(0.322185354626) / ln(2)
x=-1,6340371790199...

an other way is
f(x)=2^(3x)+3*2^x-1
f(-2)=1/64+3/4-1=-15/64 <0
f(-1)=1/8+1-1=1/8>0
==> there is a solution betheen -2<x<-1

6 0

3 years ago

HELP ME ?

julsineya [31]

1. $\sqrt{11}$ is less than 3.5

2. $\sqrt{5}$ is more than 2.1

3. -5.3 is more than -5 $\frac{3}{5}$

4. $\frac{8}{10}$ is less than 0.8 repeating

7 0

3 years ago

4. A small high school holds its graduation ceremony in the gym. Because of seating constraints, students are limited to a maxim

Ad libitum [116K]

Answer:

(a) The mean and standard deviation of X is 2.6 and 1.2 respectively.

(b) The mean and standard deviation of T are 390 and 180 respectively.

(c) The distribution of T is N (390, 180²). The probability that all students’ requests can be accommodated is 0.7291.

Step-by-step explanation:

(a)

The random variable X is defined as the number of tickets requested by a randomly selected graduating student.

The probability distribution of the number of tickets wanted by the students for the graduation ceremony is as follows:

X P (X)

0 0.05

1 0.15

2 0.25

3 0.25

4 0.30

The formula to compute the mean is:

$\mu=\sum x\cdot P(X)$

Compute the mean number of tickets requested by a student as follows:

$\mu=\sum x\cdot P(X)\\=(0\times 0.05)+(1\times 0.15)+(2\times 0.25)+(3\times 0.25)+(4\times 0.30)\\=2.6$

The formula of standard deviation of the number of tickets requested by a student as follows:

$\sigma=\sqrt{E(X^{2})-\mu^{2}}$

Compute the standard deviation as follows:

$\sigma=\sqrt{E(X^{2})-\mu^{2}}\\=\sqrt{[(0^{2}\times 0.05)+(1^{2}\times 0.15)+(2^{2}\times 0.25)+(3^{2}\times 0.25)+(4^{2}\times 0.30)]-(2.6)^{2}}\\=\sqrt{1.44}\\=1.2$

Thus, the mean and standard deviation of X is 2.6 and 1.2 respectively.

(b)

The random variable T is defined as the total number of tickets requested by the 150 students graduating this year.

That is, T = 150 X

Compute the mean of T as follows:

$\mu=E(T)\\=E(150\cdot X)\\=150\times E(X)\\=150\times 2.6\\=390$

Compute the standard deviation of T as follows:

$\sigma=SD(T)\\=SD(150\cdot X)\\=\sqrt{V(150\cdot X)}\\=\sqrt{150^{2}}\times SD(X)\\=150\times 1.2\\=180$

Thus, the mean and standard deviation of T are 390 and 180 respectively.

(c)

The maximum number of seats at the gym is, 500.

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Here T = total number of seats requested.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{T}=n\times \mu_{X}=390$