Someone pls help me I will make you as brain

PLEASE HELP WILL MARK BRAINLIEST

Nataliya [291]

Answer:

You may or may not need to include the units.

A = 18x - 18

P = 6x + 6

Graph is attached below. (2, 18)

Step-by-step explanation:

Substitute the information we need, "l" and "w", into the formulas.

l is for length, 6cm.

w is for width, (3x - 3)cm.

Use the formula for area of a rectangle.

A = lw

A = (6)(3x-3)cm²

A = (18x - 18)cm² or 18x - 18

Use the formula for perimeter of a rectangle.

P = 2(l + w)

P = 2(6 + (3x - 3))cm

P = 2(3x + 3)cm

P = (6x + 6)cm or 6x + 6

Linear equations are written in the form y = mx + b, so we do not need to factor or further simplify the formulas.

To graph, first turn the "m" value into a fraction form.

8 -> 8/1

6 -> 6/1

You need two points to graph each line.

For each equation, the first point is on the y-axis at the "b" value. Then use the "m" in the equation to count the number of units up (numerator) and to the right (denominator).

The solution is (2,18)

8 0

3 years ago

Solve 2(b − 3) ≥ 3(3 − b) and describe the graph of the solution. Show all your work.

solong [7]

Answer:

$\large\boxed{b\geq3\to b\in[3,\ \infty)}$

Step-by-step explanation:

$2(b-3)\geq3(3-b)\qquad\text{use the distributive property:}\ a(b+c)=ab+ac\\\\(2)(b)+(2)(-3)\geq(3)(3)+(3)(-b)\\\\2b-6\geq9-3b\qquad\text{add 6 to both sides}\\\\2b-6+6\geq9+6-3b\\\\2b\geq15-3b\qquad\text{add}\ 3b\ \text{to both sides}\\\\2b+3b\geq15-3b+3b\\\\5b\geq15\qquad\text{divide both sides by 5}\\\\\dfrac{5b}{5}\geq\dfrac{15}{5}\\\\b\geq3$

$,\ \geq-\text{line to the right}\\-\text{o}\text{pen circle}\\\leq,\ \geq-\text{closed circle}$

4 0

3 years ago

Simplify the fraction completly<br> 60/100

nignag [31]

Answer: 3/5

Step-by-step explanation: simplified completely

6 0

3 years ago

Read 2 more answers

ICE Princess25 [194]

Answer:

Yes, the triangles are simirar with a scale factor of 3/2.

Step-by-step explanation:

We find the scale factor by dividing corresponding sides:

12/8=3/2

18/12=3/2

24/16=3/2

4 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

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

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago