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3 years ago

Which of the following are measurements of the sides of a right triangle?

2 answers:

6 0

By using pythagoras theoram
h^2=a^2+b^2
26^2=24^2+10^2
676=576+100
676=676
26, 24, 10 are measurements of the sides of a right triangle

Gekata [30.6K]3 years ago

5 0

Answer:

D) all of the above

Step-by-step explanation:

Since, By the Pythagoras theorem,

In a right angle the sum of the square of two small sides is equal too the square of the large side.

Since,

$10^2=8^2+6^2$

⇒ They are measures of a right triangle.

$13^2=12^2+5^2$

⇒ They are measures of a right triangle.

$26^2=24^2+10^2$

⇒ They are measures of a right triangle.

Hence, option 'D' is correct.

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The area under a particular normal curve between 6 and 8 is 0.695. A normally distributed variable has the same mean and standar

TiliK225 [7]

Answer:

69.5%

Step-by-step explanation:

A feature of the normal distribution is that this is completely determined by its mean and standard deviation, therefore, if two normal curves have the same mean and standard deviation we can be sure that they are the same normal curve. Then, the probability of getting a value of the normally distributed variable between 6 and 8 is 0.695. In practice we can say that if we get a large sample of observations of the variable, then, the percentage of all possible observations of the variable that lie between 6 and 8 is 100(0.695)% = 69.5%.

4 0

3 years ago

I need help- ;-; Please help me-

dalvyx [7]

Graph A
it shows two points for each x, meaning its not a function.

7 0

2 years ago

Read 2 more answers

Given that p = 3, q= 5 and r= 7, work out the value of 2pqr.

mario62 [17]

Answer:

$2pqr =2 (3)(5)(7) = 210$

3 0

2 years ago

Read 2 more answers

If u do 3 I’ll mark Brainlynest u don’t have to do 3 it depends but also 14 points

Helen [10]

Answer:

Step-by-step explanation:

A(-5,-2),B(-2,-2), C(-5,2)

find distance first:d=√(x2-x1)^2+(y^2-y1^2)

AB: 3

BC: 5

AC: 4

perimeter of triangle=3+5+4=12

compliment angle=90

90+34=124

5x+17+x+19=180

x=24

4 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,

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

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{T}=n\times \sigma_{X}=180$

So, the distribution of T is N (390, 180²).

Compute the probability that all students’ requests can be accommodated, i.e. less than 500 seats were requested as follows:

$P(T$

Thus, the probability that all students’ requests can be accommodated is 0.7291.

8 0

3 years ago