A line's y-intercept is -10, and its slope is 1. What is its equation in slope-intercept form?

Bryce tried to solve an equation step by step. \qquad\begin{aligned} \dfrac83&=3\left(c+\dfrac53\right)\\\\ \\ \dfrac83&

sesenic [268]

Answer:

Bryce is wrong in step 1 because he did not distribute 3 over 5/3

Explanation

Given the steps taken by bryce as shown, we are to find where he made an error

$\qquad\begin{aligned} \dfrac83&=3\left(c+\dfrac53\right)\\\\ \\ \dfrac83&=3c+\dfrac53&\green{\text{Step } 1}\\\\ \\ 1&=3c&\blue{\text{Step } 2}\\\\ \\ \dfrac13&=c&\purple{\text{Step } 3}\\\\ \end{aligned}$

Given the expression;

$\dfrac83&=3\left(c+\dfrac53\right)\\\\ \\$

Step 1:Expand the bracket using the distributive law;

8/3 = 3c + 3(5/3)

Simplify

8/3 = 3c + 15/3

Step 2: Subtract 15/3 from both sides

8/3 - 15/3 = 3c+15/3-15/3

(8-15)/3 = 3c

-7/3 = 3c

Step 3: Multiply both sides by 1/3

-7/3 * 1/3 = 3c * 1/3

-7/9 = c

Swap

c = -7/9

From the calculation, we can see that Bryce is wrong in step 1 because he did not distribute 3 over 5/3 thereby making his solution incorrect

3 0

3 years ago

I toss an unfair coin 12 times. This coin is 65% likely to show up heads. Calculate the probability of the following.

mars1129 [50]

Answer:

a. 0.0368

b. 0.99992131

c. 0.2039

d. 0.0048

e. 0.6533

Step-by-step explanation:

Let the probability of obtaining a head be p = 65% = 13/20 = 0.65. The probability of not obtaining a head is q = 1 - p = 1 -13/20 = 7/20 = 0.35

Since this is a binomial probability, we use a binomial probability.

a. The probability of obtaining 11 heads is ¹²C₁₁p¹¹q¹ = 12 × (0.65)¹¹(0.35) = 0.0368

b. Probability of 2 or more heads P(x ≥ 2) is

P(x ≥ 2) = 1 - P(x ≤ 1)

Now P(x ≤ 1) = P(0) + P(1)

= ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹

= (0.65)⁰(0.35)¹² + 12(0.65)¹(0.35)¹¹

= 0.000003379 + 0.00007531

= 0.0007869

P(x ≥ 2) = 1 - P(x ≤ 1)

= 1 - 0.00007869

= 0.99992131

c. The probability of obtaining 7 heads is ¹²C₇p⁷q⁵ = 792(0.65)⁷(0.35)⁵ = 0.2039

d. The probability of obtaining 7 heads is ¹²C₉q⁹p³ = 220(0.65)³(0.35)⁹ = 0.0048

e. Probability of 8 heads or less P(x ≤ 8) = ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹ + ¹²C₂p²q¹⁰ + ¹²C₃p³q⁹ + ¹²C₄p⁴q⁸ + ¹²C₅p⁵q⁷ + ¹²C₆p⁶q⁶ + ¹²C₇p⁷q⁵ + ¹²C₈p⁸q⁴

= = ¹²C₀(0.65)⁰(0.35)¹² + ¹²C₁(0.65)¹(0.35)¹¹ + ¹²C₂(0.65)²(0.35)¹⁰ + ¹²C₃(0.65)³(0.35)⁹ + ¹²C₄(0.65)⁴(0.35)⁸ + ¹²C₅(0.65)⁵(0.35)⁷ + ¹²C₆(0.65)⁶(0.35)⁶ + ¹²C₇(0.65)⁷(0.35)⁵ + ¹²C₈(0.65)⁸(0.35)⁴

= 0.000003379 + 0.00007531 + 0.0007692 + 0.004762 + 0.01990 + 0.05912 + 0.1281 + 0.2039 + 0.2367

= 0.6533

3 0

3 years ago

I need help on 7 and 8 plssssss

lilavasa [31]

I think the domain and range are infinite

3 0

3 years ago

Read 2 more answers

How many 5-member chess teams can be chosen from 15 interested players? Consider only the members selected, not their board posi

ryzh [129]

Answer:

3003 number of 5-member chess teams can be chosen from 15 interested players.

Step-by-step explanation:

Given:

Number of the interested players = 15

To Find:

Number of 5-member chess teams that can be chosen = ?

Solution:

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula $nCr = \frac{n!}{r!(n - r)!}$

where

n represents the total number of items,

r represents the number of items being chosen at a time.

Now we have n = 15 and r = 5

Substituting the values,

$15C_5 = \frac{15!}{5!(15- 5)!}$

$15C_5 = \frac{15!}{5!(10)!}$

$15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 }{5 \times 4 \times 3 \times 2 \times 1(10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}$