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vlabodo [156]
3 years ago
11

Justin wants to rent a boat and spend at most 36. The boat costs 6 per hour, and Justin has a discount coupon for 4 off. What ar

e the possible numbers of hours Justin could rent the boat?
Mathematics
1 answer:
Maru [420]3 years ago
7 0

Answer:

The possible numbers of hours Justin could rent the boat (0, 40/6]

Step-by-step explanation:

Let t represents the number of hours the boat is used,

∵ The boat costs 6 per hour,

So, the original cost of boat = 6t

Now, after getting $ 4 off

Final cost = 6t - 4

Now, final cost ≤ $ 36

\implies 6t - 4 \leq 36

6t\leq 36 + 4

6t \leq 40

t\leq \frac{40}{6}

Also, number of hours can not be negative

⇒ t ≥ 0

But, he must use the boat ⇒ t ≠ 0

Hence, the possible numbers of hours Justin could rent the boat,

(0, \frac{40}{6}]

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A village wishes to measure the quantity of water that is piped to a factory during a typical morning. A gauge on the water line
anzhelika [568]

Answer:

570 m³

Step-by-step explanation:

The volume of water is the product of flow rate of water and the time taken. We are to get the volume of water used between 6 am and 9 am, that is for 3 hours (9 - 6).

We are given the flow rate at 6 am and the flow rate at 9 am, but this flow rate changes between 6 am and 9 am. To get the estimate of the water used, Let us assume that it flows at the same flow rate as it was at 6 am throughout, hence:

V_L= 100 m^3/h * 3h=300\ m^3

Also, let us assume that it flows at the same flow rate as it was at 9 am throughout, hence:

V_R= 280 m^3/h * 3h=840\ m^3

To get the best estimate of the total volume, let us find the average of the two values, hence:

Volume=\frac{V_L+V_R}{2} =\frac{300+840}{2}=570\ m^3

3 0
2 years ago
If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95​% confident that the d
katen-ka-za [31]

Answer:

We need a sample size of least 119

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

Sample size needed

At least n, in which n is found when M = 0.09

We don't know the proportion, so we use \pi = 0.5, which is when we would need the largest sample size.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.09 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.09\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.09}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.09})^{2}

n = 118.6

Rounding up

We need a sample size of least 119

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3 years ago
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Answer:

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Step-by-step explanation:

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