All categories

2 years ago

How do I make both these equations into a elimination equation? Look at the pic I’m very confused!

Mathematics

1 answer:

qwelly [4]2 years ago

6 0

Answer:

Step-by-step explanation:

c + a = 342 ..................... (1)

5c + 12a = 2550 ........... (2)

- 5 × (1) + (2)

7a = 840

a = 120

c = 342 - 120 = 222

You might be interested in

Let f(x) = -2x - 7 and g(x) = -4x + 6. Find (fxg)(-5). a. -59 b. 3 c. 26 d. -6

DedPeter [7]

Answer:

Step-by-step explanation:

The given functions are:

$f(x) = - 2x - 7$

and

$g(x) = - 4x + 6$

$(f \times g)(x) = f(x) \times g(x)$

$(f \times g)(x) = ( - 2x - 7)( - 4x + 6)$

When we plug in x=-5, we get:

$(f \times g)( - 5) = ( - 4 \times - 5 + 6)( - 2 \times - 5 - 7)$

$(f \times g)( - 5) = ( 20 + 6)( 10 - 7)$

$(f \times g)(5) = ( 26)( 3) =7 8$

3 0

2 years ago

Bridget rolls a fair dice 84 times. How many times would Bridget expect to roll an odd number?

cluponka [151]

Answer:

42 times

Step-by-step explanation:

Since there are 6 sides on a fair dice, each having a number 1 thru 6 with no numbers repeating and half those numbers being odd, Bridget can expect to roll an odd number 84*(1/2) times. So our answer is 42.

Also, if you could please mark this "brainliest" (the crown) it would help a lot!

6 0

2 years ago

Find the missing exponent. Set up an equation and solve. : x^4·x^n = x^5

algol [13]

Remember
(x^m)(x^n)=x^(m+n)
and
if a^m=a^n, where a=a then n=m

(x^4)(x^n)=x^5
(x^4)(x^n)=x^(4+n)
x^(4+n)=x^5
therefor
4+n=5
minus 4
n=1

5 0

2 years ago

Read 2 more answers

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:

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

See attachment (1) for histogram

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:

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

See attachment (2) for histogram

6 0

2 years ago

Anna only wants to go on 5 rides at the fair. She can either buy a pass for

lara [203]

C because 15$ is less than 30$ I have steps in comments

3 0

2 years ago