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swat32
3 years ago
12

Mason teaches ice skating. He earns

Mathematics
1 answer:
Lostsunrise [7]3 years ago
5 0

Answer:

$735.00

Step-by-step explanation:

$24.50 • 6 =147

$147 •5 = $735

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A bicycle is now selling for $315 it was discounted by 60 percent. What was the original cost.
Dovator [93]
189$
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3 0
4 years ago
I need to verify identity functions for a one to one function.
Eddi Din [679]

We have the function

f(x)=(x+6)^3

1. For f^-1:

Let y = f(x) = (x+6)^3

Switch x and y to get:

x=(y+6)^3

And solve for y

\begin{gathered} x^{\frac{1}{3}}=y+6 \\ x^{\frac{1}{3}}-6=y+6-6 \\ x^{\frac{1}{3}}-6=y \end{gathered}

And we have y = f^-1(x)

Answer blank 1:

f^{-1}(x)=x^{\frac{1}{3}}-6

2. For f o f^-1 (x):

(f\circ f^{-1})(x)=f(f^{-1}(x))

And solve

\begin{gathered} =f(x^{\frac{1}{3}}-6) \\ =(x^{\frac{1}{3}}-6+6)^3 \\ =(x^{\frac{1}{3}})^3 \\ =x \end{gathered}

answer blank 2

x^{\frac{1}{3}}-6

answer blank 3

x^{\frac{1}{3}}-6

answer blank 4

x^{\frac{1}{3}}

3. For f^-1 o f:

(f^{-1}\circ f)(x)=f^{-1}(f(x))

Solve

\begin{gathered} =f^{-1}((x+6)^3) \\ =\sqrt[3]{(x+6)^3}-6 \\ =x+6-6 \\ =x \end{gathered}

answer blank 5

(x+6)^3

answer blank 6

(x+6)^3

answer blank 7

x+6

4 0
1 year ago
Factor z2 – 3z – 18.     
SVETLANKA909090 [29]
Answer is b !!!!!!!!!!!!!!
8 0
4 years ago
Read 2 more answers
Hope Amelia Solo, the American soccer goalkeeper, World Cup champion and two-time Olympic gold medalist, allows goals at a rate
Nuetrik [128]

Answer:

3.70% probability she allows more than one goal

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

X~Pois(0.3)

This means that \mu = 0.3

In her next match, what is the probability she allows more than one goal?

Either she allows at most one goal, or she allows more than one goal. The sum of the probabilities of these events is decimal 1. So

P(X \leq 1) + P(X > 1) = 1

We want P(X > 1). So

P(X > 1) = 1 - P(X \leq 1)

In which

P(X \leq 1) = P(X = 0) + P(X = 1)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.3}*(0.3)^{0}}{(0)!} = 0.7408

P(X = 1) = \frac{e^{-0.3}*(0.3)^{1}}{(1)!} = 0.2222

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.7408 + 0.2222 = 0.9630

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.9630 = 0.0370

3.70% probability she allows more than one goal

3 0
3 years ago
I do be needing help tho ngl
Sliva [168]

Answer:

6^5

Step-by-step explanation:

( I put ^5 to show that I mean 6 with an exponent of 5) Hope this helped

3 0
3 years ago
Read 2 more answers
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