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3 years ago

9

See the attached image for the question

1 answer:

Tpy6a [65]3 years ago

5 0

Answer:

2

Step-by-step explanation:

7/5 + 3/5 so 2 is the answer

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There are blue, red and yellow marbles in a bag. The probability of choosing a blue marble is 1/8, and the probability of choosi

Ilya [14]

Answer:

3/8

Step-by-step explanation:

So what I did is make 1/2 have a common denominator with 1/8. The common denominator would be 4/8. Then i added 1/8 and 4/8 and got 5/8. Then I subtracted 5/8 and 8/8 because 8/8 would be the 100% chance. Then I got 3/8. So I would then think the chance of getting a yellow marble would be 3/8 chance. I hope I helped :)

3 0

3 years ago

Find the value of 7xº - (6x)".

docker41 [41]

7x - 1 + 6x - 1 = 180
13x - 2 = 180
13x = 182
x = 14

6 0

3 years ago

Read 2 more answers

Enter the plan width for the rectangular room. Round your answer to the nearest tenth.

Jobisdone [24]

Answer:

350

Step-by-step explanation:

5x7x10=350

5 0

3 years ago

2. The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability th

coldgirl [10]

Answer:

0.30

Step-by-step explanation:

Probability of stopping at first signal = 0.36 ;

P(stop 1) = P(x) = 0.36

Probability of stopping at second signal = 0.54;

P(stop 2) = P(y) = 0.54

Probability of stopping at atleast one of the two signals:

P(x U y) = 0.6

Stopping at both signals :

P(xny) = p(x) + p(y) - p(xUy)

P(xny) = 0.36 + 0.54 - 0.6

P(xny) = 0.3

Stopping at x but not y

P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06

Stopping at y but not x

P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24

Probability of stopping at exactly 1 signal :

P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30

8 0

3 years ago

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information: $P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000$ , where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

$\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2$

Equating it to zero we get,

$x^2 + 36x - 6400 = 0$

We use the quadratic formula to find the values of x:

$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ , where a, b and c are coefficients of $x^2, x^1 , x^0$ respectively.

Putting these value we get x = -100, 64

Now, again differentiating

$\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x$

At x = 64, $\displaystyle\frac{d^2(P(x))}{dx^2} < 0$

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0

3 years ago