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3 years ago

I really don’t understand what the questions asking? How do I solve this

Mathematics

1 answer:

Eddi Din [679]3 years ago

5 0

Answer:

$17.85

Step-by-step explanation:

Let the dollar amount per shirt be represented by s.

Let the dollar amount per pants be represented by p.

You have the following system:

3s+2p=104.81

2s+1p=61.33

If I multiply the bottom equation by -2, I can set the system up for elimination since the equations are already and the same form and I will have 2p and -2p in the same column there. 2p+(-2p)=0

Multiplying equation 2 by -2:

3s+2p=104.81

-4s-2p=-122.66

-----------------------Add the equations:

-1s =-17.85

Multiply both sides by -1:

s=17.85

So one shirt cost $17.85

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Derrick had $100 to spend at a sporting goods store. He bought a pair of shorts for $11.36 and a pair of athletic shoes for $47.

olga_2 [115]

Answer:

$40.82

Step-by-step explanation:

Subtract $11.36 from $100 and you'll get $88.64 and then just subtract $47.82 and the answer is $40.82. Not sure if u know how to subtract but just line the decimals up and take the smaller number away from the bigger one so in this case it would just be 100-11.36=88.64 and since he spent more money it would be 88.64-47.82=40.82

6 0

2 years ago

Read 2 more answers

The points (4,k) and (5,-8) fall on a line with a slope of -7. what is the value of k?

babunello [35]

Answer:

k= -1

Step-by-step explanation:

To find k, use the points and the m =-7 in the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}\\\\-7=\frac{-8-k}{5-4} \\\\-7 = \frac{-8-k}{1}\\\\-7=-8-k\\\\k = -8+7\\\\k=-1$

4 0

3 years ago

Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let X(t) be the number of custom

mash [69]

Answer:

Step-by-step explanation:

Given that:

X(t) = be the number of customers that have arrived up to time t.

$W_1,W_2$ ... = the successive arrival times of the customers.

(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

$E(W_!|X(t)=2) = \int\limits^t_0 {X} ( \dfrac{d}{dx}P(X(s) \geq 1 |X(t) =2))$

$= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}$

$= 1 - \dfrac{P(X(s) \leq 0 , 1 \leq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

$= 1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

Now $P(X(s) \leq 0) = P(X(s) = 0)$

(b) We can Determine the conditional mean E[W3|X(t)=5] as follows;

$E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\ = 1- P (X(s) \leq 2 | X (t) = 5 ) \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}$