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Aloiza [94]
3 years ago
6

Someone help plz, I'm stuck

Mathematics
1 answer:
VladimirAG [237]3 years ago
6 0

Answer:

Original price reduced by 30%

Step-by-step explanation:

You want the deepest discount, which means you want the greatest percent/amount off. The first choice is 20%. The second is 30%. The third one is 25%, because you are paying for 75% (100-75). The second choice offers the deepest discount because it is 30% off.

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Simplify the expression. Write the answer using scientific notation. (9 x 10)^2(2 x 10)^10
Valentin [98]

Answer:

B

Step-by-step explanation:

To multiply scientific notation, multiply each part of the number.

(9 x 10)^2(2 x 10)^10 = (9*2) * (10^ 2 + 10) = 18 * 10 ^ 12

Since scientific notation can only have a digit in the ones place from 1-9. 18*10^12 becomes 1.8*10^13.

The solution is B.

7 0
3 years ago
The three circles are all centered at the center of the board and are of radii 1, 2, and 3, respectively.Darts landing within th
FromTheMoon [43]

Answer:

a) P=0.262

b) P=0.349

c) P=0.215

d) E(x)=12.217

e) P=0.603

Step-by-step explanation:

The question is incomplete.

Complete question: "The three circles are all centered at the center of the board (square of side 6) and are of radii 1, 2, and 3, respectively.Darts landing within the circle of radius 1 score 30 points, those landing outside this circle, butwithin the circle of radius 2, are worth 20 points, and those landing outside the circle of radius2, but within the circle of radius 3, are worth 10 points. Darts that do not land within the circleof radius 3 do not score any points. Assume that each dart that you throw will land on a point uniformly distributed in the square, find the probabilities of the accompanying events"

(a) You score 20 on a throw of the dart.

(b) You score at least 20 on a throw of a dart.

(c) You score 0 on a throw of a dart.

(d) The expected value on a throw of a dart.

(e) Both of your first two throws score at least 10.

(f) Your total score after two throws is 30.

As the probabilities are uniformly distributed within the area of the board, the probabilities are proportional to the area occupied by the segment.

(a) To score 20 in one throw, the probabilities are

P(x=20)=P(x=20\&30)-P(x=30)=\frac{\pi r_2^2}{L^2} -\frac{\pi r_1^2}{L^2}\\\\P(x=20)=\frac{\pi(r_2^2-r_1^2)}{L^2}=\frac{3.14*(2^2-1^2)}{6^2}=\frac{3.14*3}{36} =0.262

(b) To score at least 20 in one throw, the probabilities are:

P(x\geq 20)=P(x=20\&30)=\frac{\pi r_2^2}{L^2}=\frac{3.14*2^2}{6^2} =0.349

(c) To score 0 in one throw, the probabilities are:

P(x=0)=1-P(x>0)=1-\frac{\pi r_3^2}{L^2} =1-\frac{3.14*3^2}{6^2} =1-0.785=0.215

(d) Expected value

E(x)=P(0)*0+P(10)*10+P(20)*20+P(30)*30\\\\E(x)=0+\frac{\pi(r_3^2-r_2^2)}{L^2}*10+ \frac{\pi(r_2^2-r_1^2)}{L^2}*20+\frac{\pi(r_1^2)}{L^2}*30\\\\E(x)=\pi[\frac{(3^2-2^2)}{6^2}*10+\frac{(2^2-1^2)}{6^2}*20+\frac{1^2}{6^2}*30]\\\\E(x)=\pi[1.389+1.667+0.833]=3.889\pi=12.217

(e) Both of the first throws score at least 10:

P(x_1\geq 10; x_2\geq 10)=P(x\geq 10)^2=(\frac{\pi r_3^2}{L^2} )^2=(\frac{3.14*3^2}{6^2} )^2=0.785^2=0.616

(f) Your total score after two throws is 30.

This can happen as:

1- 1st score: 30, 2nd score: 0.

2- 1st score: 0, 2nd score: 30.

3- 1st score: 10, 2nd score: 20.

4- 1st score: 20, 2nd score: 10.

1 and 2 have the same probability, as do 3 and 4, so we can add them.

P(2x=30)=2*P(x_1=30;x_2=0)+2*P(x_1=20;x_2=10)\\\\P(2x=30)=2*P(x_1=30)P(x_2=0)+2*P(x_1=20)P(x_2=10)\\\\P(2x=30)=2*\frac{\pi r_1^2}{L^2}*(1-\frac{\pi r_3^2}{L^2})+2*\frac{\pi(r_2^2-r_1^2)}{L^2}*\frac{\pi(r_3^2-r_2^2)}{L^2}\\\\P(2x=30)=2*\frac{\pi*1^2}{6^2}*(1-\frac{\pi*3^2}{6^2})+2*\frac{\pi(2^2-1^2)}{6^2}*\frac{\pi(3^2-2^2)}{6^2}\\\\P(2x=30)=2*0.872*0.215+2*0.262*0.436=0.375+0.228=0.603

6 0
3 years ago
The box plots compare the number of calories in each snack pack of crackers and cookies.
Nikitich [7]

Answer:

4th statement is true.

Step-by-step explanation:

We have been two box plots, which represents the number of calories in each snack pack of crackers and cookies. We are asked to find the correct statement about our given box plots.

1. More packets of crackers have 80 calories than any other number of calories.

We can see that median of box plot representing calories of cookies is 80. This means that half of the packets of crackers have less than 80 calories and half of the packets have more than 80 calories, therefore, 1st statement is false.

2. The value 70 is an outlier for the number of calories in the cookie pack.

Since an outlier is 1.5 times the interquartile range.

IQR=Q_3-Q_1

\text{IQR of cookie packs}=105-90

\text{IQR of cookie packs}=15

\text{Lower outlier}=Q_1-(1.5*IQR)

\text{Lower outlier}=90-(1.5*15)

\text{Lower outlier}=90-22.5

\text{Lower outlier}=67.5

Since any number less than 67.5 will be an outlier and 70 is grater than 67.5, therefore, 70 is not an outlier in number of calories in cookie packs and 2nd statement is false.

3. The upper quartile of the cookie data is equivalent to the maximum in the cracker data.

We can see that upper quartile of cookie data is 105 and the maximum in cracker data is 100. Since 105 is greater than 100, therefore, 3rd statement is false.

4. The number of calories in each pack of cookies has a greater variation than the number of calories in each pack of crackers.

Since range and IQR are good measures of variation of box-plots, so we will find the range and IQR of our both box-plots.

We have already seen that IQR of cookie packs is 15.

\text{IQR of cracker packs}=85-75

\text{IQR of cracker packs}=10

\text{Range}=\text{Maximum value - Minimum value}

\text{Range of calories in cracker packs}=100-70

\text{Range of calories in cracker packs}=30

\text{Range of calories in cookie packs}=115-70

\text{Range of calories in cookie packs}=45  

We can see that the range of calories in cookie packs (45) is greater than range of calories in cracker packs (30) and IQR of calories in cookie packs (15) is greater than IQR of calories in cracker packs (10), therefore, 4th statement is true.

3 0
4 years ago
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Use a scientific calculator to evaluate ln 26 (round to the nearest ten-thousandth).
tamaranim1 [39]
Ln 26 = 3.2581 to the nearest ten thousandth
3 0
4 years ago
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Solve the compound inequalities. <br> 3x &lt; 12 or 3x - 6 &gt; 18
gtnhenbr [62]
3x<12 divide that by 3x on both sides, x will equal 4 so it will be x<4
4 0
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