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1 year ago

15

It costs $3 to bowl a game and $2 for shoe rental. a. Write an expression for the total cost (in dollars) of $g$ games. Expressi

on: b. Use your expression to find the total cost of 8 games. Cost of 8 games: $

1 answer:

Semmy [17]1 year ago

7 0

Answer: a. Total cost = 3g+2

b. The total cost of 8 games is $26.

Step-by-step explanation:

Given: Cost to bowl a game = $3

Shoe rental charge = $2 (One -time charge)

Total cost = (Cost to bowl a game) x (Number of games)+ (Shoe rental charge )

Let g= number of games, then the expression to find the total cost will be :

Total cost = 3g+2

To find the total cost of 8 games, put g=8, we get

Total cost = 3(8)+2

= 24+2 = $26

Therefore , the total cost of 8 games is $26.

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Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

1 year ago

Given g(x) = x2 + 3x - 19, find g(-2)

BlackZzzverrR [31]

Answer:

-4 + -6 -19

-10 -19

-29

5 0

2 years ago

Lenmana started studying how the number of branches on her tree grows over time. The number of branches increases by a factor of

Ad libitum [116K]

Answer:

$N(t) = 48 * 11/6^{\frac{t}{3.5} }$

Step-by-step explanation:

The formula for <u>exponential growth</u> is y = ab^x.

To write this equation, we know it has to start with 48 (which is the variable a). We need to add the rate of growth. This is 11/6 (which is variable b). But we also need to account for the "every 3.5 years" part, so divide the x as an exponent by 3.5.

N(t) = 48 * 11/6^(t/3.5)

This equation is easy to test, and it's a good idea to test it after you write it. For example, after 3.5 years we know that it should have 48*11/6 branches. Does our equation work? Yes.

7 0

1 year ago

Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well

laiz [17]

Answer:

The value is $P(A) = 0.133617$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 7.5$

The standard deviation is $\sigma = 0.2$

The safest water level is between 7.2 and 7.8

Generally the probability that the selected pool has a pH level that is not considered safe is mathematically represented as

$P(A) = 1 - P(7.2 \le X \le 7.8 )$

Here

$P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - \mu }{\sigma } < \frac{X - \mu }{ \sigma }$

Generally $\frac{X - \mu }{ \sigma } = Z (The \ standardized \ value \ of X )$

So

$P(7.2 < X < 7.8 ) = P(\frac{ 7.2 - 7.5 }{0.2 } < Z$

$P(7.2 < X < 7.8 ) = P(-1.5 < Z$

=> $P(7.2 < X < 7.8 ) = P(Z < 1.5) - P( Z < - 1.5)$

From the z-table the probability of (Z < -1.5) and ( Z <1.5) are

$P(Z < 1.5) = 0.93319$

and

$P(Z < -1.5) = 0.066807$

So

$P(7.2 < X < 7.8 ) =0.93319 - 0.066807$

$P(7.2 < X < 7.8 ) =0.0866383$

So

$P(A) = 1 - 0.0866383$

=> $P(A) = 0.133617$

5 0

1 year ago

You found $6.60 on the ground at school, all in nickels, dimes, and quarters. you have twice as many quarters as dimes and 42 co

Veseljchak [2.6K]

Let the number of nickels found be n, the number of dimes be d and the number of quarters be q.

i) "you have twice as many quarters as dimes and 42 coins in all."

means that 2d=q, and n+d+q=42

we can reduce the number of unknowns by substituting q with 2d:

n+d+q=42
n+d+2d=42
n+3d=42

We can write all n, d and q in terms of d as follows:

there are n=42-3d nickels, d dimes and q=2d quarters.

ii) In total there are $6.60 dollars,

1 nickel = 5 cent = $0.05
1 dime = 10 cent = $0.1
1 quarter = 25 cent = $0.25

thus

(42-3d)*0.05 + d*0.1 +2d*0.25= $6.60

2.1 - 0.15d+0.1d+0.5d=6.60

2.1+0.45d=6.6

0.45d=6.6-2.1=4.5

d=4.5/0.45=10

iii)

so, there are 10 dimes, 2d=2*10=20 quarters and 42-3d=42-3*10=12 nickels.

Answer: 10 dimes, 20 quarters, 12 nickels

8 0

1 year ago