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3 years ago

14

1 liter of petrol cost r13.6 how many liter r200?

1 answer:

gulaghasi [49]3 years ago

6 0

The answer I would say is 14

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Im so lost, please help me asap

almond37 [142]

Answer:

it's so simple ....

x + 141 = 135

x = 135 - 141

x = -6

it means ur answer is a

8 0

2 years ago

Assume that the mean length of a human pregnancy is 269 days. If 95% of all human pregnancies last between 249 and 289 days, w

o-na [289]

Answer:93% not sure

Step-by-step explanation:

7 0

2 years ago

Q. The probability that a student passes a statistics test is 2/3 and the

kenny6666 [7]

Answer:

4/9 or 44.44%

Step-by-step explanation:

We have the following probabilities:

Probability of winning the statistical test = P (s) = 2/3

Probability of winning the mathematics and statistics test = P (s n m) = 14/45

Probability of winning at least one test = P (s u m) = 4/5

We have that, the probability of winning the mathematics test P (m) can be found in the following way:

P (s n m) = P (s) + P (m) - P (s u m)

replacing we have:

14/45 = 2/3 + P (m) - 4/5

P (m) = 14/45 - 2/3 + 4/5

p (m) = 4/9 = 0.444

Which is the same as there is a 44.44% probability that I will win the math test

5 0

3 years ago

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

3 0

3 years ago

Hello there >3 i would like explanation for this :)

Blizzard [7]

Answer:

75

Step-by-step explanation:

Let t and u represent the tens and units digits, respectively.

t = u + 2 . . . . the tens digit exceeds the units digit by 2

t + 2u = 17 . . . . the sum of the tens digit and twice the units digit is 17

Substituting for t in the second equation, we get ...

(u +2) +2u = 17

3u = 15 . . . . . . . . subtract 2

u = 5 . . . . . . . . . . divide by 3

t = u+2 = 7

The number is 75.

_____

<em>Check</em>

7 exceeds 5 by 2. 7 + twice 5 = 7 + 10 = 17.

8 0

3 years ago