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4 years ago

What is the value of sin(t) if cos(t)=0.1, and t is in quadrant 1

Mathematics

1 answer:

Marizza181 [45]4 years ago

8 0

Answer:

$sin(t)=\frac{\sqrt{99}}{10}$

Step-by-step explanation:

Remember that

If angle t belong to the First Quadrant

then

The value of sin(t) and cos(t) are positive values

we know that

$sin^2(t)+cos^2(t)=1$ ---> by trigonometric identity

we have

$cos(t)=0.1=\frac{1}{10}$

substitute

$sin^2(t)+(\frac{1}{10})^2=1$

$sin^2(t)+\frac{1}{100}=1$

$sin^2(t)=1-\frac{1}{100}$

$sin^2(t)=\frac{99}{100}$

$sin(t)=\frac{\sqrt{99}}{10}$

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Suppose 72 gallons of water came out of a pipe in 8 minutes.

Masteriza [31]

Answer:

9 gallons of water per minute

Step-by-step explanation:

First identify what you know:

1) The pipe was flowing water for eight minutes.

2) In that eight minutes, 72 gallons of water came out.

So, if 72 gallons of water came out in eight minutes, how many gallons flowed out in a single minute? Well, assuming that the flow rate was consistent, then simply divide the total number of gallons that flowed out, by the number of minutes the pipe was open!

72/8 = 9

The pipe released 9 gallons of water per minute, thus resulting in 72 gallons of water in 8 minutes!

Hope this helps! :)

4 0

3 years ago

Customers arrive at a grocery store at an average of 2.1 per minute. Assume that the number of arrivals in a minute follows the

Black_prince [1.1K]

Answer:

a) P(2)=0.270

b) P(X>3)=0.605

c) P=0.410

Step-by-step explanation:

We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the Poisson distribution:

$\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}$

a) In this case: $\lambda=2.1$

$P(2)=\frac{2.1^2 \cdot e^{-2.1}}{2}\\\\P(2)=0.270$

Therefore, the probability is P(2)=0.270.

b) In this case: $\lambda=2\cdot 2.1=4.2$

$P(X>3)=1-P(X\leq 3)\\\\P(X>3)=1-\sum_{x=0}^3 \frac{4.2^x \cdot e^{-4.2}}{x!}\\\\P(X>3)=1-0.395\\\\P(X>3)=0.605$

Therefore, the probability is P(X>3)=0.605.

c) We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.

In this case: $\lambda=2\cdot 2.1=4.2$

$P(X\geq 5)=1-P(X$