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3 years ago

Need help with question 3. Anything is appreciated thank you:)

Mathematics

1 answer:

Paladinen [302]3 years ago

7 0

Hey boo! The answer to three is a, and b. Have a fantastic day! Love your handwriting btw!

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On 12/31/18, Hite Industries reported retained earnings of $475,000 on its balance sheet, and it reported that it had $135,000 o

olganol [36]

Answer:

Dividends = $105000

So option (c) will be correct answer

Step-by-step explanation:

We have given that Retained earning on 12/31/18 is $475000

And retained earning on 12/31/18 is $445000

Net income = $135000

Change in retained income = $475000-$445000 = $30000

We have to find dividends

We know that dividends is given by

Dividends = net income - change in retained income = $135000-$30000 = $105000

So option (c) will be correct answer

7 0

3 years ago

I WILL MARK BRAINLIEST FOR CORRECT ANSWERS

e-lub [12.9K]

Answer:

Relation 1 and 2 are not functions

Step-by-step explanation:

7 0

3 years ago

Let X be the number of the cars being repaired at a repair shop. We have the following information:

worty [1.4K]

Answer:

(a) Sample Space

$S = \{0,1,2,3\}$

(b) PMF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

(c) CDF

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

Step-by-step explanation:

Solving (a): The sample space

From the question, we understand that at most 3 cars will be repaired.

This implies that, the number of cars will be 0, 1, 2 or 3

So, the sample space is:

$S = \{0,1,2,3\}$

Solving (b): The PMF

From the question, we have:

$P(2) = P(1)$

$P(0) = P(3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$

$P(1\ or\ 2) = 0.5 * P(0\ or\ 3)$ can be represented as:

$P(1) + P(2) = 0.5[P(0) + P(3)]$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(1) + P(1) = 0.5[P(0) + P(0)]$

$2P(1) = 0.5[2P(0)]$

$2P(1) = P(0)$

$P(0)= 2P(1)$

Also note that:

$P(0) + P(1) + P(2) + P(3) = 1$

Substitute $P(2) = P(1)$ and $P(0) = P(3)$

$P(0) + P(1) + P(1) + P(0) = 1$

$2P(1) + 2P(0) = 1$

Substitute $P(0)= 2P(1)$

$2P(1) + 2*2P(1) = 1$

$2P(1) + 4P(1) = 1$

$6P(1) = 1$

Solve for P(1)

$P(1) = \frac{1}{6}$

To calculate others, we have:

$P(2) = P(1)$

$P(2) = P(1) = \frac{1}{6}$

$P(0)= 2P(1)$

$P(0) =2 * \frac{1}{6}$ $P(0) =\frac{1}{3}$

$P(3) = P(0) =\frac{1}{3}$

Hence, the PMF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {P(x)} & {1/3} & {1/6} & {1/6} & {1/3} \ \end{array}$

<em>See attachment (1) for histogram</em>

Solving (c): The CDF ; F(x)

This is calculated as:

$F(x) = P(X \le x) =\sum\limit^{3}_{x_i \le x} P(x_i)$

For x = 0;

We have:

$P(X \le 0) = P(0)$

$P(X \le 0) = 1/3$

For x = 1

$P(X \le 1) = P(0) + P(1)$

$P(X \le 1) = 1/3 + 1/6$

$P(X \le 1) = 1/2$

For x = 2

$P(X \le 2) = P(0) + P(1) + P(2)$

$P(X \le 2) = 1/3 + 1/6 + 1/6$

$P(X \le 2) = 2/3$

For x = 3

$P(X \le 3) = P(0) + P(1) + P(2) + P(3)$

$P(X \le 3) = 1/3 + 1/6 + 1/6 + 1/3$

$P(X \le 3) = 1$

Hence, the CDF is:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ {F(x)} & {1/3} & {1/2} & {2/3} & {1} \ \end{array}$

<em>See attachment (2) for histogram</em>

6 0

3 years ago

Simplify completely quantity 2x to the power of 2 + 16x + 30

Tems11 [23]

Answer:

the answer to this equation is c

2(x+5) over 5x-2

6 0

3 years ago

Arithmetic sequences algebra

andreyandreev [35.5K]

As given arithmetic sequence is :-

-17 , -13 , -9

So, here we have to find a12

a ( 1st term ) = -17

d ( common difference) = a2 - a1

d = -13 -(-17)

d = 4

nth place = a12

Now we know that to find any term we apply formula :-

an = a + (n-1)d

So putting Values in above mentioned formula :-

a12 = -17 + ( 12-1) 4

a12 = -17 + (11)4

a12 = -17 + 44

Hope it helps

3 0

3 years ago