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Kryger [21]
3 years ago
13

Lisa needed 1 cup of white sugar to 2 cups of brown sugar to make one batch of cookies.What would be the ratio of white sugar to

brown sugar if she made 4 batches of cookies?Explain your reasoning.
Mathematics
1 answer:
raketka [301]3 years ago
4 0
1 cup : 2 cups of brown sugar

1:2

4 batches of cookies, so multiply both numbers by 4. 

4:8

Answer: 4 to 8
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Suppose the average tread-life of a certain brand of tire is 42,000 miles and that this mileage follows the exponential probabil
Ahat [919]

Answer: The probability that a randomly selected tire will have a tread-life of less than 65,000 miles is 0.7872 .

Step-by-step explanation:

The cumulative distribution function for exponential distribution is :-

P(X\leq x)=1-e^{\frac{-x}{\lambda}}, where \lambda is the mean of the distribution.

As per given , we have

Average tread-life of a certain brand of tire :  \lambda=\text{42,000 miles }

Now , the probability that a randomly selected tire will have a tread-life of less than 65,000 miles will be :

P(X\leq 65000)=1-e^{\frac{-65000}{42000}}\\\\=1-e^{-1.54761}\\\\=1-0.212755853158\\\\=0.787244146842\approx0.7872

Hence , the probability that a randomly selected tire will have a tread-life of less than 65,000 miles is 0.7872 .

8 0
3 years ago
What is the length of the arc intercepted by an angle of 10 degrees on a circle with a radius of 10 meters?
soldier1979 [14.2K]
   
\displaystyle\\
\texttt{Length of the arc }= 2\pi R\times  \frac{10}{360} = \frac{2\pi \times 10}{36} = \frac{20\pi}{36} =  \boxed{\frac{5\pi}{9}~m}



3 0
3 years ago
plot the reflection of point A across the x-axis. what are the coordinates of the reflection of point A across the x-axis? what
dsp73
You reflect point A over the x-axis, double the distance from point a to the x-axis, then give the coordinates.

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8 0
3 years ago
Y-intercept:8x +9y=40
Leto [7]

Answer:

40/9

Step-by-step explanation:

9y = -8x + 40

y = -8/9x + 40/9

5 0
3 years ago
Read 2 more answers
A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha
Mariulka [41]

Answer:

V = 63π / 200  m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

                                 y = √(42*x - x^2)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

                           S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx

- The derivative of the function f'(x) is as follows:

                            f'(x) = \frac{21-x}{\sqrt{42x-x^2} }

- The square of derivative of f(x) is:

                            f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }

- Now use the surface area formula:

                           S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi

- The Volume of the pain coating is:

                           V = S*t

                           V = 210*π*3/2000

                          V = 63π / 200 m^3

8 0
3 years ago
Read 2 more answers
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