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Liono4ka [1.6K]

2 years ago

5

A newly introduced fitness program is gaining popularity among the members of a fitness club. The number of members, s(1), who j

oined the
program t months after it was introduced is modeled by the function below.
s(t) = 10(2)* + 10
Which graph represents this function?

1 answer:

ahrayia [7]2 years ago

6 0

Answer:

Option A will be the answer.

Step-by-step explanation:

The number of members, s(t), who joined the program t months after it was introduced is modeled by the function below

$s(t) = 10(2)^{x} + 10$

The function is an exponential function and y-intercept of the function is at (0,20) point.

Now, from the graph given in the options, the first graph matches the criteria.

Hence, option A will be the answer. (Answer)

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A contaminant is leaking into a lake at a rate of R(t) = 1400e0.06t gallons/h. Enzymes have been added to the lake that neutrali

ruslelena [56]

Answer:

Step-by-step explanation:

Rate of leakage, R(t) = 1400 e^0.06t gallons/h

fraction remains , S(t) = e^(-0.32t)

initial contaminant = 1000 gallon

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G(t) = S(t) R(t)

$G(t) = 1400 e^{0.06t}\times e^{-0.32t}$

$G(t) = 1400 e^{- 0.26t}$

Put t = 18 hours

$G(t) = 1400 e^{- 0.26\times 18}$

Taking log on both the sides

ln G = ln 1400 - 0.26 x 18

ln G = 7.244 - 4.68

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5 0

2 years ago

What is 1.532 rounded to the nearest hundredth

Anni [7]

Answer:

1.53 would be the answer

Step-by-step explanation:

If you are rounding to the nearest hundredth you are rounding to the 2nd decimal place. Since the next number is under 5, we would just drop it.

8 0

1 year ago

Taylor bought 6 tickets to a movie. The price of each ticket was the same. After he bought the tickets, his account balance show

Sladkaya [172]

It is -11.34

-$68.04 divide by 6 is -$11.34, so instead of 11.34, it's -11.34

3 0

1 year ago

David proved the Law of Cosines with reference to ∆ABC using a different method. Arrange the steps of his proof in the correct s

AURORKA [14]

Answer: We can arrange the steps with help of below explanation.

Step-by-step explanation:

Here ABC is a triangle,

Draw a perpendicular from BD to side AC ( construction)

where $D\in AC$

In the right triangle BCD, from the definition of cosine:

cos C =CD/ BC

CD= a cos C

Subtracting this from the side b, we see that

DA= b-acos C

In the triangle BCD, from the definition of sine:

sin C =BD / a

BD = a sin C

In the triangle ADB, applying the Pythagorean Theorem

$c^2 =BD^2 +DA^2$

Substituting for BD and DA from (2) and (3)

$c^2 =(asinC)^2 +(b-acosC)^2$

⇒ $c^2 =a^2sin ^2C+b^2-2ab cosC+a^2 cos ^2C$ ( On simplification)

⇒ $c^2 =a^2sin ^2C+a^2 cos ^2C+b^2-2abcos C$

⇒ $c^2=a^2(sin^2C+cos^2C) +b^2-2abcosC$

⇒ $c^2 = a^2+b^2-2abcosC$ (because, $sin^2\theta+cos^2\theta =1$ )

3 0

2 years ago

Two professors are applying for grants. Professor Jane has a probability of 0.61 of being funded. Professor Joe has probability

Anna11 [10]

Answer:

a. 0.1647;
b. 0.7153;
c. 0.4453.

Step-by-step explanation:

By definition, if two event $A$ and $B$ are independent, then

$P(A\cap B) = P(A) \cdot P(B)$ . ( $P(A\cap B)$ is the probability that the outcome of both event $A$ and event $B$ are true.)

<h3>a.</h3>

Since the outcome of these two events are independent,

$\begin{aligned}P(\texttt{Jane} \cap \texttt{Joe}) &= P(\texttt{Jane}) \cdot P(\texttt{Joe})\\ &= 0.61 \times 0.27 \\&= 0.1647\end{aligned}$ .

<h3>b.</h3>

The logic not operator $\lnot$ or the prime superscript $^{\prime}$ denotes that an event does not happen.

$P(\texttt{Jane}^{\prime}) = 1 - P(\texttt{Jane}) = 1- 0.61 = 0.39$ .

$P(\texttt{Joe}^{\prime}) = 1 - P(\texttt{Joe}) = 1- 0.27 = 0.73$ .

Since the two events $\texttt{Jane}$ and $\texttt{Joe}$ , $\texttt{Jane}^{\prime}$ and $\texttt{Joe}^{\prime}$ are also independent. Probability that neither professor got funded:

$P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = P(\texttt{Jane}^{\prime}) \cdot P(\texttt{Joe}^{\prime}) = 0.39 \times 0.73 = 0.2847$ .

Probability that at least one professor got funded- in other words, it is not true that neither professor got funded:

$P((\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime})^{\prime}) = 1- P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = 0.7153$ .

<h3>c.</h3>

Similarly, since the two events $\texttt{Jane}$ and $\texttt{Joe}$ , $\texttt{Jane}$ and $\texttt{Joe}^{\prime}$ are also independent. Probability that Jane but not Joe got funded:

$P(\texttt{Jane} \cap (\texttt{Joe}^{\prime})) = P(\texttt{Jane}) \cdot P(\texttt{Joe}^{\prime}) = 0.61 \times 0.73 = 0.4453$ .

4 0

1 year ago