All categories

3 years ago

Product of -28 and sum of 3

Mathematics

2 answers:

drek231 [11]3 years ago

8 0

Answer:
-4 and 7

Explanation:
-4 x 7 = -28
-4 + 7 = 3

ale4655 [162]3 years ago

3 0

-4 and 7 would be the answer

You might be interested in

What is the value of x if... 9x - 4 - 4x = 8 + 3x

Ierofanga [76]

Hello!

Answer:

x=6

Step-by-step explanation:

Hope this helps!

7 0

2 years ago

Complete the proof to show that ABCD is a parallelogram.

Zolol [24]

Can you try to put it more in order?

8 0

3 years ago

Si tienes un rectángulo de área igual a x2 + 5x¿Cuál de las siguientes factorizaciones nos presenta el producto de la base por l

Softa [21]

Answer: factorizar x de x ^ 2 + 5x.

x (x + 5)

Step-by-step explanation:

7 0

3 years ago

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$

Since there are no replacement, the probability of taking a second non 75-W bulb is now $\frac{8}{14}=0.5714$

Following this procedure 5 times, we find the probabilities

$\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}$

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

3 0

3 years ago

If the maximum acceleration that is tolerable for passengers in a subway train is 1.74 m/s2 and subway stations are located 930

Free_Kalibri [48]

Given that the subway stations are 930 m apart, the train have to be accerelated for half the distance and then decerelated for the rest of the distance.

Recall that the distance travelled by an object with an initial velocity, u, for a period of time, t, at an accereration, a, is given by
$s=ut+ \frac{1}{2} at^2$

But, we assume that the train accelerates from rest, thus
$s=\frac{1}{2} at^2 \\ \\ \Rightarrow465=\frac{1}{2}(1.74)t^2 \\ \\ \Rightarrow t^2=534.48 \\ \\ \Rightarrow t=\sqrt{534.48}=23.12$

The maximum speed is attained at half the center of the distance between subway stations (i.e. at distance = 465 m).

Thus, maximum speed = distance / time = 465 / 23.12 = 20.11 m/s.

6 0

3 years ago