NEED HELP PLEASEEEEEEEEEEE

Use the Distributive Property to write 2(-5+3) as an equivalent expression.

Schach [20]

The correct answer is B: -4

5 0

3 years ago

A(x-b)+bx-c=a+b, when a DOES NOT = -b ; FIND X PLEASE!!! WILL GIVE BRAINLIEST!!

tangare [24]

9514 1404 393

Answer:

x = (ab +a +b +c)/(a +b)

Step-by-step explanation:

Eliminate parentheses, subtract left terms not containing x, divide by the coefficient of x.

$a(x-b)+bx-c=a+b\\\\ax+bx-ab-c=a+b\\\\x(a+b)=ab+a+b+c\\\\\boxed{x=\dfrac{ab+a+b+c}{a+b}}$

6 0

3 years ago

Prove that 1+cosθ+sinθ/1+cosθ-sinθ = 1+sinθ/cosθ

marusya05 [52]

Step-by-step explanation:

(1 + cos θ + sin θ) / (1 + cos θ − sin θ)

Multiply by the reciprocal:

(1 + cos θ + sin θ) / (1 + cos θ − sin θ) × (1 + cos θ + sin θ) / (1 + cos θ + sin θ)

(1 + cos θ + sin θ)² / [ (1 + cos θ − sin θ) (1 + cos θ + sin θ) ]

(1 + cos θ + sin θ)² / [ (1 + cos θ)² − sin² θ) ]

Distribute and simplify:

(1 + cos θ + sin θ)² / (1 + 2 cos θ + cos² θ − sin² θ)

[ 1 + 2 (cos θ + sin θ) + (cos θ + sin θ)² ] / (1 + 2 cos θ + cos² θ − sin² θ)

(1 + 2 cos θ + 2 sin θ + cos² θ + 2 sin θ cos θ + sin² θ) / (1 + 2 cos θ + cos² θ − sin² θ)

Use Pythagorean identity:

(2 + 2 cos θ + 2 sin θ + 2 sin θ cos θ) / (sin² θ + cos² θ + 2 cos θ + cos² θ − sin² θ)

(2 + 2 cos θ + 2 sin θ + 2 sin θ cos θ) / (2 cos² θ + 2 cos θ)

(1 + cos θ + sin θ + sin θ cos θ) / (cos² θ + cos θ)

Factor:

(1 + cos θ + sin θ (1 + cos θ)) / (cos θ (1 + cos θ))

(1 + cos θ)(1 + sin θ) / (cos θ (1 + cos θ))

(1 + sin θ) / cos θ

3 0

3 years ago

The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in

Stolb23 [73]

Answer:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=34.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=300-1=299$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that $t_{\alpha/2}=1.653$