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Liono4ka [1.6K]
3 years ago
13

A wildlife photographer photographs a crocodile and then paddles against a river current for 6 km. Then, hoping to photograph a

hippopotamus, she paddles 15 km across a lake. She paddles across the lake for one hour longer than she paddled up the river. Find the speed of the boat as she paddles the lake if the river current is 2 km/h.
Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

5 or 6

Step-by-step explanation:

So I did it in a different way than shown in rsm classes. I said that the speed was x-2 and x.

then I said the distance was 6 and 15. If the time taken was 1 hour more, then we can come up with this equation:

Since the time is 6/x-2 or 15/x

15/x-6/x-2=1

multiply everything to get a common denominator then cancel it out to get

(15x-30)-6x=x^2-2x

11x-30=x^2

x^2-11x+30=0

(x-5)(x-6) is factored form.

so x=5 or x=6. this means that the speed of the boat is either 5km/h or 6 km/h

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Read 2 more answers
There are two major cell phone providers in the Colorado Springs, Colorado area, one called HTC and the other, Mountain Communic
Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

Hello!

The objective of this exercise is to compare the "Churn rate" between the two major cell phone providers in colorado springs.

For each provider, the number of subscribers was noted at the beginning and end of a month.

The study variables are:

X₁: Number of customers that cut ties with HTC over a month.

X₂: Number of customers that cut ties with Mountain Communications over a month.

To compare the churn rate of both companies, the parameters of interest are the population proportions of customers that stopped using the companies services over the given period.

The hypotheses of interest are:

H₀: p₁ = p₂

H₁: p₁ ≠ p₂

α: 0.01

The statistic to use is the standard normal approximation:

Z= \frac{('p_1-'p_2)- (p_1-p_2)}{\sqrt{'p(1 - 'p)[\frac{1}{n_1} + \frac{1}{n_2}  ]} } }

Z≈N(0;1)

HTC

n₁= 10000 customers (beginning of the month)

x₁= 190 customers (people that stopped using the company service)

sample proportion 'p₁= x₁/n₁= 190/10000= 0.019

Mountain Communications

n₂= 12285 customers (beginning of the month)

x₂= 215 customers (people that stopped using the company service)

sample proportion 'p₂= x₂/n₂= 215/12285= 0.0175

pooled sample proportion 'p= \frac{x_1 + x_2}{n_1 + n_2} = \frac{190 + 215}{10000 + 12285}

'p= 0.0182

Z_{H_0}= \frac{(0.019 - 0.0175) - 0}{\sqrt{(0.0182*0.9818)[\frac{1}{10000} + \frac{1}{12285}  ]} }

Z_{H_0}= 0.833

The p-value for this test is: 0 .4048

Note: Using the p-value method to decide over a hypothesis test, the decision rule is:

If p-value ≥ α, then you do not reject the null hypothesis.

If p-value < α, then you reject the null hypothesis.

The p-value < α, so the decision is to reject the null hypothesis.

With a significance level of 1%, there is enough evidence to conclude that there is a difference between the churn rate of the two cell phone providers.

I hope it helps!

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