3.2 + (-1.8) What would that be

loris [4]

Consider a right triangle in which one of the angles $\theta$ satisfies

$\sin\theta=\dfrac35\implies\theta=\sin^{-1}\dfrac35$

That is, the side opposite $\theta$ occurs in a ratio of 3 to 5 with the hypotenuse. The side adjacent to $\theta$ then occurs in a ratio of 4 to 5 with the hypotenuse. In other words,

$\cos\theta=\dfrac45$

because

$\sin^2\theta+\cos^2\theta=\dfrac9{25}+\dfrac{16}{25}=1$

Then in this triangle,

$\cot\theta=\cot\left(\sin^{-1}\dfrac35\right)=\dfrac{\cos\theta}{\sin\theta}=\dfrac{\frac45}{\frac35}=\dfrac43$

7 0

3 years ago

How do I solve this? Solve for x 2/3+5/3x=9

ElenaW [278]

Answer:

5

Step-by-step explanation:

3 0

3 years ago

-8(6+X) + 4(2x + 7) = 0

Talja [164]

Answer:

-20 = 0

Step-by-step explanation:

-8 (6 + x) + 4 (2x + 7) = 0

-48 - 8x + 8x + 28 = 0

-20 = 0

3 0

3 years ago

using the coordinate plane shown below 1/3 point is plotted five units to the right of -4, for where should a four point be plan

Mrac [35]

we know that

the fourth point is plotted 5 units to the right of (-4,4)

so

the coordinate x of the fourth point is equal to

-4+5=1

the y-coordinate of the fourth point is the same y coordinate of (-4,4)

therefore

the coordinates of the fourth point is (1,4)

8 0

2 years ago

Suppose that the amount of time T a customer spends in a bank is exponentially distributed with an average of 10 minutes. What i

Bad White [126]

Answer:

The probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes is 0.6065.

Step-by-step explanation:

The random variable T is defined as the amount of time a customer spends in a bank.

The random variable T is exponentially distributed.

The probability density function of a an exponential random variable is:

$f(x)=\lambda e^{-\lambda x};\ x>0$

The average time a customer spends in a bank is β = 10 minutes.

Then the parameter of the distribution is:

$\lambda=\frac{1}{\beta}=\frac{1}{10}=0.10$

An exponential distribution has a memory-less property, i.e the future probabilities are not affected by any past data.

That is, P (X > s + x | X > s) = P (X > x)

So the probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes is:

P (X > 15 | X > 10) = P (X > 5)

$\int\limits^{\infty}_{5} {f(x)} \, dx =\int\limits^{\infty}_{5} {\lambda e^{-\lambda x}} \, dx\\=\int\limits^{\infty}_{5} {0.10 e^{-0.10 x}} \, dx\\=0.10\int\limits^{\infty}_{5} {e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10}|^{\infty}_{5}\\=[-e^{-0.10 \times \infty}+e^{-0.10 \times 5}]\\=0.6065$

Thus, the probability that a customer will spend more than 15 minutes total in the bank, given that the customer has already waited over 10 minutes is 0.6065.

8 0

3 years ago