Simplify the expression. Please help, really stuck!!

Explain how you can determine the number of real number solutions of a system of equations in which one equation is linear and t

DerKrebs [107]

Isolate one variable in the system of equations. Use substitution to create a one-variable equation. Then, set the quadratic equation equal to zero and find the discriminant. If the discriminant is negative, then there are no real number solutions. If the discriminant is zero, then there is one real number solution. If the discriminant is positive, then there are two real number solutions.

4 0

3 years ago

Read 2 more answers

Name three values of x that would result in y = cot x being undefined.

tester [92]

Three values of x that would result in y = cot x being undefined are $0,\pi ,2\pi$

Given :

y=cot(x)

we need to find where cot(x) is undefined

We know that $cot(x)=\frac{cotx}{sinx}$

when the denominator is 0 then we can say cot(x) is undefined

Lets find out x values that makes sin(x) as 0

We know that sin(0)=0

$sin(\pi )=0\\sin(2\pi )=0$

x values values that makes sin(x)=0 are $0,\pi ,2\pi$

These x values makes the denominator of cot(x) as 0 and it is undefined

So , three values of x that would result in y = cot x being undefined are $0,\pi ,2\pi$

Learn more : brainly.ph/question/12155214

6 0

3 years ago

A chemist has 500 ml of a 30% acid solution. She adds x milliliters of a 10% acid solution. Which statement is true about the gr

prohojiy [21]

Answer: The horizontal asymptote y = 10 means that the final concentration will always be greater than 10%.

3 0

4 years ago

You measure 50 textbooks' weights, and find they have a mean weight of 37 ounces. Assume the population standard deviation is 5.

Hoochie [10]

Answer:

90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

Step-by-step explanation:

We are given that you measure 50 textbooks' weights, and find they have a mean weight of 37 ounces.

Assume the population standard deviation is 5.2 ounces.

Firstly, the Pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weight = 37 ounces

$\sigma$ = population standard deviation = 5.2 ounces

n = sample of textbooks = 50

$\mu$ = true population mean textbook weight

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5%

level of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $37-1.645 \times {\frac{5.2}{\sqrt{50} } }$ , $37+1.645 \times {\frac{5.2}{\sqrt{50} } }$ ]

= [35.79 , 38.21]

Therefore, 90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

7 0

3 years ago

1) Find a unit rate for the following ratio. 135 words: 3 minutes

Sergio039 [100]

45. It’s just dividing 135 by 3.

5 0

3 years ago