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Wewaii [24]
2 years ago
11

Paul lives 6 miles west and 3 miles north of school. What is the direct distance from Paul's house to school?

Mathematics
2 answers:
NeTakaya2 years ago
6 0

It's (C) 6.7

a^2 + b^2 = c^2  

so 6^2 + 3^2 = c^2  

so 36 + 9 = c^2  

so 45 = c^2  

so the distance will be equal to to the square root of 45 which is about 6.7 miles.

Jobisdone [24]2 years ago
5 0
The correct answer is A) 4.5 miles. Reason being, when he walks around it is 9 miles. When he walks DIRECTLY from his house to school, it is 4.5 miles.
Hope this helps!
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artcher [175]

Answer:

AB=2.775

BC=5.55

CA=6.475

Step-by-step explanation:

Since midpoints split their sides in half, we can see that the triangle MNK formed by the midpoints will be half the perimeter of the triangle ABC. Since P of MNK = 7.4, we know that the perimeter of ABC = 7.4 * 2, which is 14.8. Now we can split the 14.8 so that it follows the ratio.

3+6+7=16

14.8/16=0.925

AB=0.925*3=2.775

BC=0.925*6=5.55

CA=0.925*7=6.475

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2 years ago
What is the operation used for the following terms: times, product, twice, etc.
Dafna1 [17]
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8 0
2 years ago
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Consider a single spin of the spinner.
alex41 [277]

Answer:

"landing on a shaded portion and landing on a 3"

"landing on an unshaded portion and landing on a number less than 2 "

Step-by-step explanation:

Mutually exclusive means the events will have no intersection.

Let's look at your first choice:

"landing on a shaded portion and landing on an even number"

Landing on a shaded portion would be 1 or 4.

Landing on an even number would be 2 or 4.

There is an intersection (they contain a common element), the 4.

These events are not mutually exclusive.

Let's look at your second choice:

"landing on a shaded portion and landing on a number greater than 3"

Landing on a shaded portion would be 1 or 4.

Landing on a number greater than 3 would be just 4.

There is an intersection; they both contain 4.

These events are not mutually exclusive.

Let's look at your third choice:

"landing on a shaded portion and landing on a 3"

Landing on a shaded portion would be 1 or 4.

Landing on 3 would just be 3.

There is no common elements in the lists listed.  These events have no intersection.

These events are mutually exclusive.

Let's look at your fourth choice:

"landing on an unshaded portion and landing on an odd number"

Landing on a unshaded portion would be 2 or 3.

Landing on an odd number would be 1 or 3.

There is an intersection; they both have 3 in common.

These events are not mutually exclusive.

Let's look at your fifth choice:

"landing on an unshaded portion and landing on a number less than 2 "

Landing on an unshaded portion would 2 or 3.

Landing on a number less than 2 would be 1.

There is no intersection.

These events are mutually exclusive.

6 0
2 years ago
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A study was performed on green sea turtles inhabiting a certain area.​ Time-depth recorders were deployed on 6 of the 76 capture
polet [3.4K]

Answer:

One can be 99% confident the true mean shell length lies within the above interval.

The population has a relative frequency distribution that is approximately normal.

Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.  

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

                    P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean shell length = 51.3 cm

             s = sample standard deviation = 6.6 cm

             n = sample of turtles = 6

             \mu = true mean shell length

Now, the 99% confidence interval for \mu =  \bar X \pm t_(_\frac{\alpha}{2}_)  \times \frac{s}{\sqrt{n} }

Here, \alpha = 1% so  (\frac{\alpha}{2}) = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> \mu  =  51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }

                                                         = [51.3 - 10.864 , 51.3 + 10.864]

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The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;

C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

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2 years ago
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