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3 years ago

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PLS HELP ME ASAP Amy must form a three-letter arrangement using only letters from the word dice. She cannot use a letter more th

an once in the arrangement. (Her arrangement doesn't need to be a valid word.)
The probability that Amy forms an arrangement that begins and ends with vowels is

The probability that Amy forms an arrangement whose letters are in alphabetical order (ascending or descending) is

2 answers:

Setler [38]3 years ago

8 0

<span>The probability that Amy forms an arrangement that begins and ends with vowels is
1/2 x 1/3 = 1/6

answer
1/6

</span><span>The probability that Amy forms an arrangement whose letters are in alphabetical order (ascending or descending) is
answer
1/3</span>

Sunny_sXe [5.5K]3 years ago

7 0

The first question is 1/5 the second is 1/26

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Which ordered pairs are in the solution set of the system of linear inequalities?

Reil [10]

Inequalities are used to represent unequal expressions

The solution to the system of inequalities is (-1,1)

<h3>How to determine the solution</h3>

The system of inequalities is given as:

y > x + 2 and

y < 2x + 3

To do this, we start by plotting the graphs of the two inequalities

y > x + 2 and y < 2x + 3

See attachment for the graph

From the attached graph, both lines meet at (-1,1)

Hence, the solution to the system of inequalities is (-1,1)

Read more about system of inequalities at:

brainly.com/question/9774970

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3 years ago

movie tickets cost $9.95 for adults and $6.75 for kids they buy 2 adult tickets and 4 students and $28 popcorn

Harrizon [31]

Answer:

the total is $153.9

Step-by-step explanation:

The total for the adults is $19.9

The children $27

Popcorn $224

$9.95×2=$19.9

$6.75×4=$27

$28×6=$168

6 0

3 years ago

A 99% confidence interval (in inches) for the mean height of a population is 65.7 < μ < 67.3. This result is based on a sa

Wittaler [7]

Answer:

$66.5 -1.976* 0.3065= 65.89$

$66.5 +1.976* 0.3065= 67.11$

And the 95% confidence interval would be givne by (65.89; 67.11)

Step-by-step explanation:

For this case we have a 99% confidence interval for the true mean hegiht givne by:

$65.7 \leq \mu \leq 67.3$

And we can begin finding the mean with this formula:

$\bar X= \frac{65.7 +67.3}{2}= 66.5$

Now we can estimate the margin of error:

$ME = \frac{67.3-65.7}{2}= 0.8$

The sample size for this case is $n = 144$ then the degrees of freedom are given by:

$df = n-1= 144-1=143$

And then we can find a critical value for a 99% of confidence and 143 degrees of freedom using a significance level of 0.01 and we got:

$t_{\alpha/2}= 2.61$

Then the standard error is given by:

$SE = \frac{ME}{t_{\alpha/2}}= \frac{0.8}{2.61}= 0.3065$

Now we can find the other critical value for 95% of confidence and we got:

$t_{\alpha/2}= 1.976$

And the new confidence interval would be given by:

$66.5 -1.976* 0.3065= 65.89$

$66.5 +1.976* 0.3065= 67.11$

And the 95% confidence interval would be givne by (65.89; 67.11)

3 0

3 years ago

A particle moves along a horizontal line. Its position function is s(t) for t is greater than or equal to 0. For each problem, f

OleMash [197]

Answer:

1) Velocity

$v(t) = -4\cdot t^{3}+36\cdot t^{2}$

Acceleration

$a(t) = -12\cdot t^{2}+72\cdot t$

2) Velocity

$v(t) = -4\cdot t^{3}+24\cdot t^{2}$

Acceleration

$a(t) = -12\cdot t^{2}+48\cdot t$

Step-by-step explanation:

From Physics we remember that velocity ( $v(t)$ ) and acceleration ( $a(t)$ ) are the first and second derivatives of the function position in time. That is:

1) Let $s(t) = -t^{4}+12\cdot t^{3}$ , where $t \ge 0$ . The functions velocity and aceleration are, respectively:

Velocity

$v(t) = -4\cdot t^{3}+36\cdot t^{2}$

Acceleration

$a(t) = -12\cdot t^{2}+72\cdot t$

2) Let $s(t) = -t^{4}+8\cdot t^{3}$ , where $t\ge 0$ . The functions velocity and acceleration are, respectively:

Velocity

$v(t) = -4\cdot t^{3}+24\cdot t^{2}$

Acceleration

$a(t) = -12\cdot t^{2}+48\cdot t$

4 0

3 years ago

Which inequality will have a shaded area below the boundary line?

sergiy2304 [10]

3x + 4y > 12 <span>will have a shaded area below the boundary line</span>

7 0

3 years ago