∆ABC transforms to produce ∆A'B'C'. Which transformation did NOT take place?

The following two sets of parametric functions both represent the same ellipse. Explain the difference between the graphs.

statuscvo [17]

The answer
ellipse main equatin is as follow:

X²/ a² + Y²/ b² =1, where a≠0 and b≠0

for the first equation: x = 3 cos t and y = 8 sin t
we can write x² = 3² cos² t and y² = 8² sin² t
and then x² /3²= cos² t and y²/8² = sin² t
therefore, x² /3²+ y²/8² = cos² t + sin² t = 1
equivalent to x² /3²+ y²/8² = 1

for the second equation, x = 3 cos 4t and y = 8 sin 4t we found
x² /3²+ y²/8² = cos² 4t + sin² 4t=1

7 0

3 years ago

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Linda cut a piece of wood for project the first cut is Shawn and can be represented by the equation Y=1/4x-5 Lisa can’t cut need

GuDViN [60]

Lisa's cut will be represented by the equation y = 1/4x + 4

6 0

3 years ago

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Freshman Algebra help please anyone (just the answer no explanation)

Vikki [24]

The answer is D. The slope is 5 and (2, 4) is on the line.

8 0

3 years ago

You are offered two different jobs for the month of July, and you decide which one to accept. Your goal is to save as much money

Monica [59]

Answer:

60 hours

Step-by-step explanation:

300 + 5x = 10x

subtract 5x from both sides to cancel 5x on the left

300 = 5x

divide both by 5

60 = x

6 0

3 years ago

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Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

Let $\mu_1$ = true mean number of sales at location A.

$\mu_2$ = true mean number of sales at location B

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, test statistics = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0

3 years ago