Massa says that 540 = 6 = 9. Is he right? Explain how you know using an equation.

Rosita and Garth are saving up for a vacation, and begin with a certain amount saved each. If Rosita works 5 hours, she will hav

noname [10]

Treat each (time, money) pair as an (x, y) pair, and get the slope of the line:
For Rosita, (5, 128), (7, 164): m = (y2 - y1)/(x2 - x1) = (164 - 128)/(7 - 5) = 18, implying that she earns $18/hr. The y-intercept is calculated as: y = 18x + b, 128 = 18*5 + b, b = $38, meaning that she started with $38. Rosita's equation is y = 18x + 38.
For Garth, (3, 124), (8, 194): m = (194 - 124)/(8 - 3) = 14. For 124 = 14*3 + b, b = $82. Garth's equation is y = 14x + 82

To find out when they will have saved the same amount, both equations would have the same y-value:
18x + 38 = 14x + 82
4x = 44
x = 11 hours
y = 18*11 + 38 = $236 (alternatively, y = 14*11 + 82 = 236)
This means that Rosita and Garth will have both saved $236 after 11 hours of working.

7 0

2 years ago

Which statement about the lines below is true?

Anika [276]

I all of those options are true

4 0

2 years ago

A random sample from a normal population is obtained, and the data are given below. Find a 90% confidence interval for . 114 157

melamori03 [73]

Answer:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.

Step-by-step explanation:

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 mean or variance 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.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

Data given: 114 157 203 257 284 299 305 344 378 410 421 450 478 480 512 533 545

The confidence interval for the population variance $\sigma^2$ is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^{17} (x_i -\bar x)^2}{n-1}}$

And in order to find the sample mean we just need to use this formula:

$\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample mean obtained on this case is $\bar x= 362.941$ and the deviation s=132.250

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=17-1=16$

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 tabel to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,16)" "=CHISQ.INV(0.95,16)". so for this case the critical values are:

$\chi^2_{\alpha/2}=26.296$

$\chi^2_{1- \alpha/2}=7.962$

And replacing into the formula for the interval we got:

$\frac{(16)(132.250)^2}{26.296} \leq \sigma \leq \frac{(16)(132.250)^2}{7.962}$

$10641.959 \leq \sigma^2 \leq 35147.074$

Now we just take square root on both sides of the interval and we got:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.

4 0

2 years ago

Anthony borrows $6.50 from a friend to pay for a sandwich and drink at lunch. Then he borrows $3.75 from his sister to pay a lib

Katen [24]

Money borrowed by Anthony from his friend for a sandwich and drink at lunch = $6.50

Money borrowed by Anthony from his sister to pay a library fine = $3.75

Hence, the equation to express Anthony's total debt that day will be =

Let T be the total debt , so the equations becomes,

6.50 + 3.75 = T

8 0

2 years ago

Read 3 more answers

Cynthia rounds a number, x,to one decimal place. the result is 6.3.

gladu [14]

The answer is: $6.25\leq6.3$

The explanation is shown below:

1. Cynthia rounds the number, which is identified as $x$ , to one decimal place and the result is 6.3.

2. Based on this, we know that $x$ could have been between 6.25 and 6.35. Therefore, the error interval for $x$ is given by:

$6.25\leq6.3$

Where $\leq$ indicates that the value 6.25 is included and indicates that the value 6.35 is not included (Because if $x$ had been exactly 6.35, Cynthia would round up to 6.4).

7 0

2 years ago