Which of the following is the solution

ANY JESUS HELPERS PLEASE HELP The incorrect work of a student to solve an equation 2(y + 4) = 4y is shown below: Step 1: 2(y + 4

wariber [46]

Answer:

2 should be distributed as 2y + 8; y = 4

Step-by-step explanation:

2(y + 4) = 4y

Distribute

2y + 8 = 4y

Step 2 is incorrect because the added instead of multiplied

Continuing on to correct the problem

Subtract 2y from each side

2y+8-2y = 4y-2y

8 = 2y

Divide by 2

8/2 = 2y/2

4 =y

3 0

3 years ago

Read 2 more answers

Pamela is making pizzas. She has 112 pieces of pepperoni. She uses 29 pieces of pepperoni on the first pizza. If she uses the sa

Svetllana [295]

Answer:

No

Step-by-step explanation:

Number of pieces of pepperoni available with Pamela = 112

Number of pieces of pepperoni used in one pizza = 29

Now, number of pepperoni left after using 29 in one pizza = 112 - 29 = 83

Now, Pamela makes second pizza, so again 29 pieces of pepperoni area used.

Now, number of pepperoni left after using 29 in one pizza = 83 - 29 = 54

Now, Pamela makes third pizza, so again 29 pieces of pepperoni area used.

Now, number of pepperoni left after using 29 in one pizza = 54 - 29 = 25

To make the third pizza, another 29 pepperoni are required but only 25 are available.

Therefore, there is not enough pepperoni for making 4 more pizza.

7 0

2 years ago

1/6 is equal to what

Anastasy [175]

1/6 = 0.1666, the last number repeating to infinity, or a repeating decimal.

Hope helps!-Aparri

6 0

3 years ago

Read 2 more answers

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$