All categories

3 years ago

Pythagorean theorem

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

4 0

Answer:

8.7

Step-by-step explanation:

hypotenuse= 14 =c

a=11

b=x

a²+b²=c²

b²=c²-a²

b²=14²-11²

b²= 196-121

b²=75

b=√75=5√3=8.7 rounded to the nearest tenth

You might be interested in

You design a tree house using a coordinate plane in which the coordinates are measured in feet. The vertices of the floor are (-

Tema [17]

Answer:

gregefdsgdsfgdfdgddg

Step-by-step explanation:

8 0

3 years ago

Is seventy five thousandths less than, greater than or equal to 0.75

ryzh [129]

Seventy five thousandths is less than .75

6 0

3 years ago

Read 2 more answers

Divide using long division. check your answers. 1. (x 3 – 13x 2 + 7x – 48) ÷ (x 2 + 3) 3. (x 3 + 5x 2 – 3x – 1) ÷ (x – 1) 5. (x

artcher [175]

1. We can conclude that the result of the division is $x-13$ with a remainder of $4x-9$ , or in other notation: $x-13+ \frac{4x-9}{x^2+3}$ . Check the steps in the first image.

3. We can conclude that the result of the division is $x^{2} +6x+3$ with a remainder of 2, or in other notation: $x^{2}+6x+3+ \frac{2}{x-1}$ . Check the steps in the firts image.

5. We can conclude that the result of the division is $x+1$ with a remainder of 5, or in other notation: $x+1+ \frac{5}{x-4}$ . Check the steps in the first image.

7. To check if $x+4$ is a factor of $x^{3}+3x^{2}-10x-24$ , we are going to perform long division. If the result of the long division has no remainder, $x+4$ is a factor of $x^{3}+3x^{2}-10x-24$ .
Since the result of the long division has no remainder, we can conclude that $x+4$ is a factor of $x^{3}+3x^{2}-10x-24$ . Check the steps in the second image.

9. To check if $x-3$ is a factor of $x^{3}+3x^{2}-10x-24$ , we are going to perform long division. If the result of the long division has no remainder, $x-4$ is a factor of $x^{3}+3x^{2}-10x-24$ .
Since the result of the long division has no remainder, we can conclude that $x-4$ is a factor of $x^{3}+3x^{2}-10x-24$ . Check the steps in the third image.

10. We can conclude that the result of the synthetic division is $-2x^{2}+9x+5$ . Check the steps in the fourth image.

3 0

3 years ago

The opponents of soccer team A are of two types: either they are a class 1 or a class 2 team. The number of goals team A scores

NikAS [45]

Answer:

a) The expected number of goals team A will score is 5.1

b) The probability that team A will score a total of 5 goals is 0.1147

Step-by-step explanation:

Let X be the amount of goals scored by team A in both matches. Let X1 and X2 be the total amount of goals team A scores in match 1 and 2 respectively, then X = X1+X2, and also

E(X) = E(X1+X2) = E(X1)+E(X2) = 0.6*2+0.4*3 + 0.3*2+0.7*3 = 5.1

b) In order for X to be equal to 5 we have 5 possibilities

- X1 is 0 and X2 is 5

- X1 is 1 and X2 is 4

- X1 is 2 and X2 is 3

- X1 is 3 and X2 is 2

- X1 is 4 and X2 is 1

- X1 is 5 and X2 is 0

Let T1 be a poisson distribution with mean λ = 2, then

$P(T1=0) = e^{-2}$

$P(T1=1) = 2 * e^{-2}$

$P(T1=2) = 2 * e^{-2}$

$P(T1=3) = \frac{4}{3} \, e^{-2}$

$P(T1=4) = \frac{2}{3}\, e^{-2}$

$P(T1=5) = \frac{4}{15}\,e^{-2}$

Lets do the same with a Poisson distribution T2 with mean λ = 3

$P(T2=0) = e^{-3}$

$P(T2=1) = 3 \, e^{-3}\\P(T2=2) = \frac{9}{2} \, e^{-3}\\P(T2=3) = \frac{9}{2} \, e^{-3}\\P(T2=4) = \frac{27}{8} \, e^{-3}\\P(T2=5) = \frac{81}{40} \, e^{-3}$

Now, we are ready to compute the probability that X is equal to 5.

$P(X1 = 0, X2 = 5) = (0.6* e^{-2} + 0.4*e^{-3}) * (0.3*\frac{4}{15}e^{-2} + 0.7*\frac{81}{40} e^{-3}) = 0.00823\\P(X1 = 1, X2 = 4) = (0.6* 2e^{-2} + 0.4*3 e^{-3}) * (0.3*\frac{2}{3}e^{-2} + 0.7*\frac{27}{8} e^{-3}) = 0.03214\\P(X1 = 2, X2 = 3) = (0.6* 2e^{-2} + 0.4*\frac{9}{2}e^{-3}) * (0.3*\frac{4}{3}e^{-2} + 0.7*\frac{9}{2} e^{-3}) = 0.05317\\P(X1 = 3, X2 = 2) = (0.6* \frac{4}{3}e^{-2} + 0.4*\frac{9}{2}*e^{-3}) * (0.3*2e^{-2} + 0.7*\frac{9}{2} e^{-3}) = 0.0471$

$P(X1 = 4, X2 = 1) = (0.6* \frac{2}{3}e^{-2} + 0.4*\frac{27}{8}e^{-3}) * (0.3*2e^{-2} + 0.7*3e^{-3}) = 0.0225\\P(X1 = 5, X2 = 0) = (0.6* \frac{4}{15}e^{-2} + 0.4*\frac{81}{40}e^{-3}) * (0.3*e^{-2} + 0.7*e^{-3}) = 0.0047$

We can conclude that

P(X = 5) = 0.00823+0.03214+0.05317+0.0471+0.0225+0.0047 = 0.1147

The probability that team A will score a total of 5 goals is 0.1147

7 0

3 years ago

Which is not a rational number

enot [183]

Answer:

Determine the type of elasticity of demand

8 0

2 years ago