All categories

2 years ago

HELP PLEASE- i attached the pic

Mathematics

2 answers:

Lerok [7]2 years ago

4 0

I think it is distributive property maybe i’m not too sure

baherus [9]2 years ago

4 0

Step-by-step explanation:

ℎ 8 2 16 ℎ . 81 8 16-8 ℎ ℎ “ “

-

You might be interested in

Write 2x + y = 17 in slope-intercept form. <br><br> Plz help me out here

Tatiana [17]

Answer:

y=-2x+17

Step-by-step explanation:

5 0

2 years ago

A sporting goods store earns 9% profit on all sales of boxing gloves. If one pair of boxing gloves sold for $43.00,what was the

Lera25 [3.4K]

The sporting goods store would earn a profit of $3.67 for a pair of boxing gloves that cost $43.00

5 0

2 years ago

What do you know to be true about the values of a and b?

Ivanshal [37]

Answer:

I don't think there is a way the values can be determined...

5 0

2 years ago

What is the inverse of the function f(x)= 1/4x-12

Tasya [4]

Answer:

The inverse is 4x+48

Step-by-step explanation:

y = 1/4x - 12

Exchange x and y

Solve for y

x = 1/4y -12

Add 12 to each side

x+12 = 1/4 y -12+12

x+12 = 1/4y

Multiply each side by 4

4(x+12) = 4*1/4y

4x+48 = y

The inverse is 4x+48

7 0

3 years ago

Read 2 more answers

If you have 6 cars, but there is only room in your driveway for 3 cars, in how many ways can you arrange the cars in your drivew

Svetradugi [14.3K]

If the order does not matter, cars can be arranged in 20 ways

If an order is important, cars can be arranged in 120 ways

The probability that the three newest cars end up parked in the driveway is 0.167

1. If the order does not matter, combination is used

$\left ({n} \atop {r}} \right.)=\frac{n!}{r!(n-r)!}$

Here, n=6 r=3

using formula,we get

$\left ( {{6} \atop {3}} \right)=\frac{6!}{3!3!} =5*4=20$

2. If an order is important, Permutation will be applicable

$^{n}P_{r} = \frac{n!}{(n-r)!} \\$

∴ $^{6} P_{3} =\frac{6!}{3!(6-3)!}=\frac{6!}{3!} =120$

3. the probability that the three newest cars end up parked in the driveway

P= $\frac{No. of possible outcomes for 3 cars}{total possibilities}$

= $\frac{6*4*5}{6!}$

= $\frac{120}{720}$

= $\frac{1}{6}$ ≈ 0.167

Hence, If the order does not matter, cars can be arranged in 20 ways

If an order is important, cars can be arranged in 120 ways

The probability that the three newest cars end up parked in the driveway is 0.167

Learn more about probability here brainly.com/question/6077878

#SPJ4

7 0

1 year ago