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2 years ago

1,000 POINTS!!!!!!!!! I NEED HELP ASAP

Mathematics

1 answer:

babunello [35]2 years ago

7 0

Answer:

x ≥ -4

Step-by-step explanation:

-3x + 2 ≤ 14

Subtract 2 from each side

-3x+2-2 ≤ 14-2

-3x ≤ 12

Divide each side by -3 . Remember that dividing by a negative flips the inequality.

-3x/-3 ≥12/-3

x ≥ -4

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An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do

vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

$P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107$

Then,

$P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007$

c) In this case, the distribution is B(1200,0.01)

$P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924$

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

$P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366$

5 0

3 years ago

ray uses a fourth of a gallon of gasoline to drive a distance of 5 1/2 miles at this rate how many miles can you drive on one ga

KIM [24]

On average you get like about 12 to 14 miles a gallon

4 0

2 years ago

Read 2 more answers

X squared plus 8X +12

Xelga [282]

Answer:

(x+2)(x+6)

Step-by-step explanation:

x2 + 8x +12

x 2

x 6

5 0

2 years ago

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

2 years ago

The number of problems that the math team successfully solved at this year's competition increased from last year by a factor of

Maksim231197 [3]

Earlier today I set you ten questions from this year’s International Singapore Maths Competition. Here are the questions and the answers. On the whole you did very well - smarter than a 10-year-old Singaporean! (With the caveat that they didn’t have multiple choice answers to choose from, and they are only ten). The only questions where your most popular answer the wrong one were 6 and 8. (C in Q6, and B in Q8). Thanks for taking part - now look through your workings...

For Year 5 pupils:

1. Mary cut off 2/5 of a piece of string. Later, she cut off another 14 m. The ratio of the length of string remaining to the total length cut off is 1 : 3. What is the length of the remaining string?

A. 5 m
B. 7 m
C. 10 m
D. 14 m
Solution is C. [73 per cent of readers got it right]

Oh Mary! This is how I would have solved it, using equations. Let L be the original length of the string, and R be what is remaining once you have cut the string twice. We know that R = (L x 3/5) – 14m, and that ((L x 2/5) +14) /R = 3, or 2L/5 + 14 = 3R. By substituting the first equation in the second we have 2L/5 + 14 = 9L/5 –42. Which rearranges to:7L/5 = 56, or L = 40. So R = 10m.

Interestingly, the Singapore method of solution is different. It requires us to think more visually about the string: We cut 2/5 of it. Then 14m, and are left with a piece that is a third of the size of what was cut. In other words, we are left with 1/4 of the original length. In order to compare the fractions 2/5 and then 1/4, lets change them to the lowest common denominator, which is 20. So, we cut off 8/20, subtract 14m and are left with 5/20. Let’s now draw the string divided into twentieths:

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The 14m must be 7/20 of the string, which mean each twentieth is 2m. The remaining piece of string is 5/20, i.e 10m

2. The areas of the faces of a rectangular box are 84 cm2, 70 cm2 and 30 cm2. What is the volume of the box?

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Photograph: ISMC
A. 300 cm3
B. 420 cm3
C. 490 cm3
D. 504 cm3
Solution is B. [85 per cent of readers got it right]

First we need to work out the possible side lengths, by seeing which two numbers multiply to be the area of each face. The 84 face could be 1 x 84, 2 x 42, 3 x 28, 4 x 21, 6 x 14 or 7 x 12. The 70 face could be 1 x 70, 2 x 35, 5 x 14 or 7 x 10. The 30 face could be 1 x 30, 2 x 15, 3 x 10 or 5 x 6.

The common factors between 84 and 70 are 1, 2, 7 and 14.
The common factors between 84 and 30 are 1, 2, 3 and 6.
The only way to make 84 with one each of these common factors are 14 from the top line and 6 from the bottom. So the edge bordering the 84 and 70 faces has length 14, and the edge bordering the 84 and 30 edges has length 6. Which means the height must be 30/6, or 70/14 = 5. Thus the volume is 14 x 6 x 5 = 420cm.

3. There are four numbers. If we leave out any one number, the average of the remaining three numbers will be 45, 60, 65 or 70. What is the average of all four numbers?

A. 50
B. 55
C. 60
D. 65
Solution is C. [82 per cent of readers got it right]

If the four numbers are A, B, C and D, then we know that

A + B + C = 45 × 3
A + B + D = 60 × 3
A + C + D = 65 × 3
B + C + D = 70 × 3
Now add them up to get 3A + 3B + 3C + 3D = (45 + 60 + 65 + 70) × 3

Which is A + B + C + D = (45 + 60 + 65 + 70) = 240. So their average is 240/4 = 60

6 0

2 years ago