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Andreyy89
3 years ago
5

In the spinner below the large wedges are twice the size of the smaller ones. What is true about the probability of landing on 6

and the probability of landing on 5?​

Mathematics
1 answer:
mr_godi [17]3 years ago
6 0

Answer:

A

Step-by-step explanation:

If we take each small section to be "1" unit, we can say the large sections (for "2" and "5") are "2" units each. So in total there will be 8 sections.

Since 5 is "2" sections, we can say:

P(5) = 2/8 = 1/4

And 6 is "1" section, so we can say:

P(6) = 1/8

Definitely, Probability of landing a 6 is HALF that of probability of landing a 5. Also we can see this is the picture.

So, from the answer choices, A is right.

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Otto invests $1,000 at age 20. Otto wants theinvestment to be worth $4,000 by age 30. Ifinterest is compounded continuously, wha
AveGali [126]
r\approx0.139

1) Since this is a Continuously Compounded operation in a 10 yrs period, then we can write out the following equation:

\begin{gathered} A=Pe^{rt} \\  \end{gathered}

2) Plugging into the equation the given data and since Otto is 20 yrs old and he plans to get $4,000 in ten years, we can write out:

\begin{gathered} 4000=1000e^{\mleft\{10r\mright\}} \\ \frac{4000}{1000}=\frac{1000e^{10r}}{1000} \\ 4=e^{10r} \\ \ln (4)=\ln (e)^{10r} \\ 10r=\ln (4) \\ r=\frac{\ln (4)}{10} \\ r=0.1386 \end{gathered}

3) Thus the rate Otto needs is

r=0.1386\approx0.139

Note that since the 0.1386 the six here is greater than 5 then we can round up to the next thousandth, in this case: 0.139. For the 0.1386 is closer to 0.139 (0.004) than to 0.138 (0.006).

Or 13.9%

8 0
1 year ago
The length of a rectangle is 5 more than the width. The perimeter is 120. What is the length, width, and area? Please HELP
Ad libitum [116K]

Answer:

l = 32.5 units, w= 27.5 units, A = 893.75 units²

Step-by-step explanation:

width is w

length is l = 5+w

P = 2( l+w) , substitute l for 5+w

P = 2(5+w+w)

P = 2(5+2w)

P = 10 +4w

P = 120

10 +4w = 120

4w = 120-10

4w = 110

w= 110/4

w= 27.5 units

l = 5+w = 5+ 27.5 = 32.5 units

A = l*w = 27.5 * 32.5 = 893.75 units ²

5 0
3 years ago
5 1/2 by 4 3/4 by 1/4 scale it down by 1/10
TiliK225 [7]

Answer:

(11/20 by 19/40 by 1/40) OR (0.55 by 0.475 by 0.025)

Step-by-step explanation:

To scale down a set of measurements, simply multiply the original measurement by the factors you're scaling it by. For example, take the first measurement of 5 1/2. To scale this down, I turned the value into an improper fraction (11/2) to make the multiplication process easier. Then, I multiplied 11/2 by the factor of 1/10.

After that, I got 11/20, which is the new first measurement.

7 0
3 years ago
37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the
Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy

and I assume <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by

D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}

• Compute the line integral directly by splitting up <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}

(b) <em>C</em> is the boundary of the region

D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}

• Compute the line integral directly, splitting up <em>C</em> into 3 components,

<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1

<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}

• Using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}

4 0
3 years ago
The graph shows the feasible region for the system with constraints:
Dimas [21]

Answer:

The vertices feasible region are (0 , 15) , (10 , 15) , (20 , 5)

The minimum value of the objective function C is 125

Step-by-step explanation:

* Lets look to the graph to answer the question

- There are 3 inequalities

# y ≤ 15 represented by horizontal line (purple line) and cut the

  y-axis at point (0 , 15)

# x + y ≤ 25 represented by a line (green line) and intersected the

  x-axis at point (25 , 0) and the y- axis at point (0 , 25)

# x + 2y ≥ 30 represented by a line (blue line) and intersected the

  x-axis at point (30 , 0) and the y-axis at point (0 , 15)

- The three lines intersect each other in three points

# The blue and purple lines intersected in point (0 , 15)

# The green and the purple lines intersected in point (10 , 15)

# The green and the blue lines intersected in point (20 , 5)

- The three lines bounded the feasible region

∴ The vertices feasible region are (0 , 15) , (10 , 15) , (20 , 5)

- To find the minimum value of the objective function C = 4x + 9y,

  substitute the three vertices of the feasible region in C and chose

  the least answer

∵ C = 4x + 9y

- Use point (0 , 15)

∴ C = 4(0) + 9(15) = 0 + 135 = 135

- Use point (10 , 15)

∴ C = 4(10) + 9(15) = 40 + 135 = 175

- Use point (20 , 5)

∴ C = 4(40) + 9(5) = 80 + 45 = 125

- From all answers the least value is 125

∴ The minimum value of the objective function C is 125

6 0
3 years ago
Read 2 more answers
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