A cargo ship traveled 150 nautical miles in 6 hours at a constant speed. How far did the cargo ship travel in one hour

Mathematics

2 answers:

mixas84 [53]2 years ago

7 0

Answer:

25 nautical miles

Step-by-step explanation:

We need to find the rate. We can do so using the formula, distance = rate * time

d= rt

We know the distance and the time, so we need to solve for the rate

d/t = rate

d= 150 nautical miles

t = 6 hours

d/t = 150 nautical miles /6 hours

rate = 25 nautical miles per hour

So in 1 hour we can go 25 nautical miles

TEA [102]2 years ago

3 0

Answer:

25 nautical miles.

Step-by-step explanation:

Givens

d = 150 nautical miles
t = 6 hours
r = ???

Equation

d = r * t

Solution

150 nautical miles = r * 6 hours Divide by 6 hours
150 nautical miles/6 hours = r * 6 hours / 6 hours
25 nautical miles / hour = r

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The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>