Write each of the equations below in standard form (i.e. y = ax^2+bx+c y=ax ​2 ​​ +bx+c):
a. y = 3(x-7)(x+4) y=3(x−7)(x+4)
b. y = 2(x-3)^2 + 5 y=2(x−3) ​2 ​​ +5
c. y = -2(x+3)(x+5) y=−2(x+3)(x+5)
d. y = -3(x-4)^2 + 8y=−3(x−4) ​2 ​​ +8

Answer:

Given : A bag contains 30 lottery balls numbered 1-30 a ball is selected replaced then another is drawn.

To find : Each probability  

1) p ( and even,then odd )  

2) p ( 7, then a number greater than 16)

3) p ( a multiple of 5, then a prime number )  

4) p ( two even number )

Solution :

[tex]\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}[/tex]

1) There are 15 even numbers and 15 odd numbers.

Probability of getting even first then odd is

[tex]\text{P(even,then odd)}=\frac{15}{30}\times\frac{15}{30}[/tex]

[tex]\text{P(even,then odd)}=\frac{225}{900}=\frac{1}{4}[/tex]

2) Number greater than 16 out of 30 are 14.

Probability of getting 7 first then a number greater than 16 is

[tex]\text{P(7, then a number greater than 16)}=\frac{1}{30}\times\frac{14}{30}[/tex]

[tex]\text{P(7, then a number greater than 16)}=\frac{14}{900}=\frac{7}{450}[/tex]

3) Multiple of 5 - 5,10,15,20,25,30=6

Prime numbers - 2,3,7,9,11,13,17,19,23,29=10

Probability of getting a multiple of 5, then a prime number  is

[tex]\text{P(a multiple of 5, then a prime number )}=\frac{6}{30}\times\frac{10}{30}[/tex]

[tex]\text{P(a multiple of 5, then a prime number )}=\frac{60}{900}=\frac{1}{15}[/tex]

4) There are 15 even numbers.

Probability of getting  two even number is

[tex]\text{P( two even number)}=\frac{15}{30}\times\frac{15}{30}[/tex]

[tex]\text{P( two even number)}=\frac{225}{900}=\frac{1}{4}[/tex]

Without the visual reference of an image, the specific geometric construction represented can not be accurately determined. The provided options relate to different types of geometric constructions. These include copying angles, constructing parallel lines, copying segments, and creating perpendiculars from a point not located on a line.

The geometric construction represented by the image is not entirely clear without an image. However, if we are to interpret the options, we can make some educated assumptions. Copy an angle construction involves replicating an existing angle in a new location. A parallel lines construction typically involves creating a line parallel to an existing one. To copy a segment construction, you would reproduce a particular line segment. The option of creating a perpendicular from a point not on the line would typically involve making a line perpendicular to an existing line from a specific point not originally on that line. Without the image, we can't definitively answer the question.

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The answer is  an = 9 • 4n - 1

The explicit rule for the [tex]nth[/tex] term of the given sequence is [tex]a_n=9(4)^{n-1}[/tex].

Given:

The given sequence is [tex]9,36,144,576[/tex].

To find:

The explicit rule for the [tex]nth[/tex] term of the given sequence.

Explanation:

the first term of the sequence is [tex]9[/tex].

The ratios of two consecutive terms are:

[tex]\dfrac{36}{9}=4[/tex]

[tex]\dfrac{144}{36}=4[/tex]

[tex]\dfrac{576}{144}=4[/tex]

The given sequence is a geometric sequence because the sequence has a common ratio [tex]4[/tex].

The explicit formula for the [tex]nth[/tex] term is:

[tex]a_n=ar^{n-1}[/tex]  

Where, [tex]a[/tex] is the first term and [tex]r[/tex] is the common ratio.

Substituting [tex]a=9,r=4[/tex], we get

[tex]a_n=9(4)^{n-1}[/tex]  

Therefore, the explicit rule for the [tex]nth[/tex] term of the given sequence is [tex]a_n=9(4)^{n-1}[/tex].

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A two-way frequency table is used to organize bivariate data into a format that helps calculate relative frequencies, empirical probabilities, and analyze marginal and conditional distributions.

A two-way frequency table allows you to organize bivariate data. This type of table is particularly useful in displaying data concerning two categorical variables, such as gender and sports preferences. The table sets up data in a way that makes it easier to calculate relative frequency and, as a result, empirical probability. It also assists in organizing the data for marginal and conditional distributions. Joint frequencies are the counts in the body of the table, while the marginal frequencies are located in the table's margins, summarizing the totals for each variable across all categories. Conditional distributions can then be analyzed, focusing on particular subsets within the table to assess probabilities within those subsets.

Answer with explanation:

 It is given that ,Marge correctly guessed whether a fair coin turned up "heads" or "tails" on sic consecutive flips.

When we flip a coin , there are two possible Outcomes, one is Head and another one is Tail , that is total of 2.

Probability of an event

         [tex]=\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}[/tex]

Probability of getting head

                      [tex]=\frac{1}{2}[/tex]

Probability of getting tail

                      [tex]=\frac{1}{2}[/tex]

⇒There can be two guesses , either it will be true and another one will be false.

So, Possible outcome of correct guess={True, False}=2

--Probability of Incorrect(False) guess

                       [tex]=\frac{1}{2}[/tex]

--Probability of Correct(True) guess in seventh toss

                       [tex]=\frac{1}{2}\\\\=\frac{1}{2} \times 100\\\\=50 \text{Percent}[/tex]

⇒Probability that she will correctly guess the outcome of the Seventh coin toss, if previous sixth tosses has correct guess

  =T×T×T×T×T×T×T, where T=True guess

  = 0.5×0.5×0.5×0.5×0.5×0.5×0.5

 =0.0078125

=0.0078 (approx)

The dimensions of the rectangle are [tex]\( \boxed{9 \text{ km} \times 12 \text{ km}} \)[/tex].

Let's denote the length of the rectangle as [tex]\( l \)[/tex] kilometers, and its width as [tex]\( w \)[/tex] kilometers.

From the problem statement, we have two pieces of information:

1. The width is 6 kilometers less than twice the length:

  [tex]\[ w = 2l - 6 \][/tex]

2. The area of the rectangle is 108 square kilometers:

  [tex]\[ lw = 108 \][/tex]

Now we can substitute the expression for \( w \) from the first equation into the second equation:

[tex]\[l(2l - 6) = 108\][/tex]

Expand and simplify the equation:

[tex]\[2l^2 - 6l = 108\][/tex]

Subtract 108 from both sides to set the equation to zero:

[tex]\[2l^2 - 6l - 108 = 0\][/tex]

Divide every term by 2 to simplify:

[tex]\[l^2 - 3l - 54 = 0\][/tex]

Now, we'll solve this quadratic equation using the quadratic formula, [tex]\( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -54 \)[/tex]:

[tex]\[l = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-54)}}{2 \cdot 1}\][/tex]

[tex]\[l = \frac{3 \pm \sqrt{9 + 216}}{2}\][/tex]

[tex]\[l = \frac{3 \pm \sqrt{225}}{2}\][/tex]

[tex]\[l = \frac{3 \pm 15}{2}\][/tex]

This gives us two possible solutions for [tex]\( l \)[/tex]:

[tex]\[l = \frac{18}{2} = 9 \quad \text{or} \quad l = \frac{-12}{2} = -6\][/tex]

Since length cannot be negative, we take [tex]\( l = 9 \)[/tex] kilometers.

Now, substitute [tex]\( l = 9 \)[/tex] back into the expression for [tex]\( w \)[/tex]:

[tex]\[w = 2l - 6 = 2 \cdot 9 - 6 = 18 - 6 = 12\][/tex]

Therefore, the dimensions of the rectangle are:

- Length [tex]\( l = 9 \)[/tex] kilometers

- Width [tex]\( w = 12 \)[/tex] kilometers

To verify, calculate the area:

[tex]\[l \times w = 9 \times 12 = 108 \text{ square kilometers}\][/tex]

Since this matches the given area, the dimensions [tex]\( l = 9 \)[/tex] kilometers and [tex]\( w = 12 \)[/tex] kilometers are correct.

Thus, the dimensions of the rectangle are [tex]\( \boxed{9 \text{ kilometers} \times 12 \text{ kilometers}} \)[/tex].

[tex] |\Omega|=52\\
|A|=26\\\\
P(A)=\dfrac{26}{52}=\dfrac{1}{2}=50\% [/tex]

Answer:

280

Step-by-step explanation:

Answer:

B. 6.3 cm

Step by step explanation:

We have been given measure of central angle which intercepts to our minor arc XY.  

Since we know that the formula to find measure of arc length is:

[tex]\text{Arc length}=\frac{\theta}{360}\times \text{circumference of circle}[/tex]

[tex]\text{Arc length}=\frac{\theta}{360}\times {2\pi r}[/tex]  

Now let us substitute our given values in above formula.

[tex]\theta=40^{o}[/tex] and [tex]radius=9 cm[/tex]

[tex]\text{Arc length}=\frac{40}{360}\times {2\pi \cdot 9}[/tex]

[tex]\text{Arc length}=\frac{1}{9}\times {2\pi \cdot 9}[/tex]

[tex]\text{Arc length}=2 \pi [/tex]

[tex]\text{Arc length}=6.2831853071795865\approx 6.3[/tex]

Therefore, length of minor arc XY is 6.3 cm and option B is the correct choice.

Final answer:

To find the total distance Mark has run after 5 days, multiply his daily distance of 3/4 mile by 5 days, resulting in A.3 3/4 miles.

Explanation:

The question asks us to find the total distance that Mark runs over 5 days, given that he runs 3/4 of a mile each day. To find the total distance, we simply multiply the daily distance by the number of days.

Multiply the daily distance (3/4 mile) by the number of days (5 days):

(3/4) × 5 = 15/4

Convert the improper fraction to a mixed number:

15/4 is equivalent to 3 whole miles and 3/4 of a mile, which can be written as 3 3/4 miles.

Therefore, after 5 days, Mark has run a total distance of A.3 3/4 miles.

The area of the rectangle at L = 14 cm and B = 9 cm is increasing at 97 cm²/s.

The area of a rectangle is given by -

A[R] = L x B

Given is the length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 5 cm/s.

Now, we can write -

dL/dt = 3 cm/s

dB/dt = 5 cm/s

We know, that the area is -

A = LB

differentiating both sides with respect to [t], we get -

dA/dt = L dB/dt + B dL/dt

dA/dt = 5L + 3B

At L = 14 cm and B = 9 cm.

(dA/dt) [14, 9] = 5 x 14 + 3 x 9 = 70 + 27 = 97 cm²/s

Therefore, the area of the rectangle at L = 14 cm and B = 9 cm is increasing at 97 cm²/s.

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Answer: 247.215 Hz

Step-by-step explanation:

We know that in music theory, a perfect fifth is a musical interval having inverse perfect fourth that corresponds a pair of pitches with a frequency ratio of 3:2.

Let the  frequency of the note a perfect fifth above [tex]E_3[/tex] be x, then we have the following proportion.

[tex]x:164.81::3:2\\\\\Rightarrow x=\dfrac{3\times164.81}{2}=247.215[/tex]

Hence, the  frequency of the note a perfect fifth above [tex]E_3[/tex] is 247.215 Hz.

[tex] |\Omega|=156\\
|A|=156-147=9\\\\
P(A)=\dfrac{9}{156}=\dfrac{3}{52}\implies \text{D} [/tex]

Answer:  

The correct answer is C. 42.2%

Step-by-step explanation:

Total number of cards in the deck of playing cards = 52

Number of spades in the deck of playing cards = 13

Number of cards other than spade = 52 - 13

                                                          = 39

If she draws a card 3 times and doesn't get a spade, she loses.

So, She loses only if he gets all the three cards other than spade

[tex]\text{Probability that she does not get a spade in first draw = }\frac{39}{52}[/tex]

Now, The card is replaced if she does not get a spade.

[tex]\text{So, Probability that she does not get a spade in second draw = }\frac{39}{52}[/tex]

[tex]\text{Similarly, Probability that she does not get a spade in third draw = }\frac{39}{52}[/tex]

[tex]\text{Thus, Probability that she will lose the game = }\frac{39^3}{52^3}=0.422[/tex]

[tex]\text{Also, The percentage of Probability that she will lose the game = }0.422\times 100=42.2\%[/tex]

Hence, The correct answer is C. 42.2%