All categories

3 years ago

I don’t understand!!!

Mathematics

1 answer:

marishachu [46]3 years ago

8 0

Answer:

Step-by-step explanation:

We can use momentum and conservation of momentum equation to solve this problem. The equation would be:

$m_1v_1+m_2v_2=(m_1+m_2)v_3$

m_1 is given as 2

v_1 is given as 3

m_2 is what we want

v_2 is 0 since they are sticking together after collision

v_3 is the final velocity at the resultant end, which is 1

Substituting and solving via algebra:

$m_1v_1+m_2v_2=(m_1+m_2)v_3\\(2)(3)+m_2(0)=(2+m_2)(1)\\6+0=2+m_2\\6=2+m_2\\m_2=6-2=4$

Thus, mass of 2nd box is 4kg

Answer choice A is right

You might be interested in

Marla did 65 set ups each day for a week use the distributive property to show an expression you can use to find the total numbe

Maslowich

<span>X = 65(7)
X = 455
Marla did 455 sit-ups in a week</span>

5 0

3 years ago

Estimate the square root of 300 and 190

mihalych1998 [28]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Pretty please lmk what this is

Gnesinka [82]

Answer:

Donda

Step-by-step explanation:

3 0

3 years ago

15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.

Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = $30C5 = \frac{30 !}{(30-5)!5!} \\$

$30C5 = 142506$ ways

Number of ways of selecting 5 good bulbs from 26 bulbs = $26C5 = \frac{26 !}{(26-5)!5!} \\$

$26C5 = 65780$ ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = $\frac{26C4 * 4C1}{30C5}$

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) = $\frac{26C3 * 4C2}{30C5}$

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) = $\frac{26C2 * 4C3}{30C5}$

Pr(3 defective) = 0.009

Pr(4 defective) = $\frac{26C1 * 4C4}{30C5}$

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0

3 years ago

What are some examples of how you have used, or could use, algebraic equations to solve problems in your everyday life?

pshichka [43]

We use algebraic equations in our daily lives without knowing it. For example, when we go shopping, calculate grocery expenses, or even fill our gas tanks, we are using equations. Let's say we needed to fill our gas tank, but we only had $15.00 to do so. If the price of gas is $3.00 per gallon, how many gallons could we buy? We use the simply algebraic equation of $\frac{x}{y}$ to calculate the total gallons that can be bought. x = the money we have. y = price per gallon of gas. $\frac{15}{3}$ = 5. So with $15.00, we can buy 5 gallons of gas; and that is just one example of using an algebraic equation in everyday life!

3 0

4 years ago