It costs $20 for 4 t shirts. Find the cost of 1 t shirt.

What is a reasonable estimate for the solution? O (1,-3/4) O (-3/4,1) O (-1,3/4) O (3/4,-1)

Zina [86]

Answer:

See Explanation

Step-by-step explanation:

Your question is incomplete, as the equations or graph or table(s) were not given.

However, I'll give a general way of solving this.

Take for instance, the equations are:

$y = \frac{4}{3}x - 1$

$y = \frac{2}{3}x - \frac{1}{2}$

To do this, we start by equating both equations.

$y = y$

i.e.

$\frac{4}{3}x - 1= \frac{2}{3}x - \frac{1}{2}$

Collect Like Terms

$\frac{4}{3}x - \frac{2}{3}x= 1 - \frac{1}{2}$

Take LCM

$\frac{4x- 2x}{3}= \frac{2 - 1}{2}$

$\frac{2x}{3}= \frac{1}{2}$

Cross Multiply

$2x * 2 = 3 * 1$

$4x = 3$

Make x the subject

$x = \frac{3}{4}$

Substitute 3/4 for x in $y = \frac{4}{3}x - 1$

$y = \frac{4}{3} * \frac{3}{4} - 1$

$y = 1 - 1$

$y = 0$

Hence:

$(x,y) = (\frac{3}{4},0)$

6 0

3 years ago

Help please!!!!!!!!!

ANEK [815]

The answer is (8,40.32)

4 0

3 years ago

Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Norm

Vlad1618 [11]

Answer:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

The distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar x =512$ represent the sample mean

$\sigma=100$ represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

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

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

$Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

$25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}$

And we can find the critical value $z_{\alpha/2}$ like this:

$z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576$

And we know that on the right tail of the z score =2.576 we have $\alpha/2$ of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998= $\alpha/2$ , so then $\alpha=0.00498*2=0.009995 \approx 0.01$ , and then we can find the confidence like this:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

8 0

3 years ago

Help pleaseeeeeeeee

Doss [256]

Answer:9

Step-by-step explanation: im not really sure I kinda just guess and check, so like, 2/3 times 3 is 6/3 which is 2. so 6/2 is 3. 3 times 3 is nine.

4 0

3 years ago

Write the quadratic equation in standard form. What is the value of b 2 - 4ac?

Fittoniya [83]

B^2 - 4ac > 0
2x^2 +7x -1 =0
b =7 a=2 c= -1
b^2 - 4ac = 7^2 - 4*2*(-1)
= 49 + 8
= 57

5 0

3 years ago