Harry was given an assignment. He had 42 days to get 336 tasks done at work.Which operation would he use to figure out how many tasks he needs to complete each day

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

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:  

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%

The circular rug with a diameter of 8 feet covers approximately 50.27 square feet of floor space. This was calculated by finding the area of the circle using the formula πr².

The question refers to how much floor space a circular rug with a diameter of 8 feet will cover. To solve this, you need to calculate the area of the circle. The formula for the area of a circle is πr², where r is the radius of the circle. But we were given the diameter, which is twice the radius. So the radius of the rug is 8/2 = 4 feet.

Substitute r=4 into the formula: Area = π(4²) = 16π square feet. Therefore, the circular rug covers 16π, or approximately 50.27 square feet of floor space.

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Answer:

280

Step-by-step explanation:

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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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)

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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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.

Using Poisson Distribution, the probability that the box will contain less than three defective bolts is computed by adding up the probabilities of finding 0, 1, or 2 defective bolts in a box. The average occurrence rate is 4 defective bolts on average.

Based on the provided details, we can conclude that this is a problem involving Poisson Distribution. The key parameters here are that the box contains 100 bolts and on average, 4 bolts are defective. The question asks for the probability that the box will contain less than three defective bolts, or in other words the probability that there are either 0, 1, or 2 defective bolts in a box. With a Poisson distribution, the formula to calculate the probability is P = lambda^k * exp(-lambda) / k! where lambda is the average occurrence rate, k is the number of occurrences of the event, and exp refers to the exponential function.

To calculate the probability of finding less than 3 defective bolts, we add together the probabilities of finding 0, 1, and 2 defective bolts using the above formula. In this example, lambda = 4 defective bolts on average, and we substitute k with 0, 1, and 2 respectively. Solve these cases separately and then add them up, that will give you the probability that a box will contain less than three defective bolts.

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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]

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.

Answer:

substituition property

Step-by-step explanation: