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3 years ago

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

Mathematics

1 answer:

Zina [86]3 years ago

6 0

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:

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

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graph of horizontal hyperbola on a coordinate plane going through 4 comma 0 and negative 4 comma 0 with points at 6 and 4 tenths

Alex_Xolod [135]

Answer:

x squared over 16 minus y squared over 25 equals 1

Step-by-step explanation:

x²/4² - y²/5² = 1

x²/16 - y²/25 = 1

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4 years ago

You are conducting a study of three types of feed supplements for cattle to test their effectiveness in producing weight gain am

maw [93]

K is the number of groups in one way ANOVA. Since here 4 groups involved. So k here is four.

<u>Explanation:</u>

In statistics, one-way analysis of variance is a technique that can be used to compare means of two or more samples. This technique can be used only for numerical response data, the "Y", usually one variable, and numerical or categorical input data, the "X", always one variable, hence "one-way".

The one-way analysis of variance (ANOVA) is used to determine whether there are any statistically significant differences between the means of two or more independent (unrelated) groups (although you tend to only see it used when there are a minimum of three, rather than two groups).

4 0

3 years ago

According to the U.S. Census Bureau, 42% of men who worked at home were college graduates. In a sample of 480 women who worked a

Likurg_2 [28]

Answer:

1) $0.327 - 0.539 \sqrt{\frac{0.327(1-0.327)}{480}}=0.315$

The point estimate for the proportion of college graduates among women who work at home is 0.327

2) $0.327 - 0.539 \sqrt{\frac{0.327(1-0.327)}{480}}=0.315$

$0.327 + 0.539 \sqrt{\frac{0.327(1-0.327)}{480}}=0.339$

The 80% confidence interval is given by (0.315; 0.339)

Step-by-step explanation:

For this case we have the following info given:

$X = 157$ represent the women who worked at home who were college graduates

$n = 480$ the sample size selected

Part 1

In order to find the proportion of college graduates among women who work at home and we can use the following formula:

$\hat p=\frac{X}{n} = \frac{157}{480}= 0.327$

The point estimate for the proportion of college graduates among women who work at home is 0.327

Part 2

Construct an 80% confidence interval for the proportion of women who work at home who are college graduates. Round the answer to three decimal places. An 80% confidence interval for the proportion of women who work at home Is < p <

The confidence interval for the true proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$