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seraphim [82]
3 years ago
15

Two ships are near a buoy in the open ocean. One ship is 20 km due north of the buoy, and the other ship is 13.5 km due east of

the buoy.
Round the answer to the nearest tenth.
Mathematics
1 answer:
anastassius [24]3 years ago
6 0

Answer:

24 km is your answer

Step-by-step explanation:

Let x be the distance between the two ships.

Hypotenuse side = x km

Adjacent side = 13.5 km

Perpendicular side = 20 km

by using Pythagoras theorem:

x²= 13.5²+ 20²=582.25

x=√ 582.25

х =24.129857024 km

Therefore, the distance between two ships to the nearest tenth place is, 24.1 km

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4 0
3 years ago
What is nine less than a number?
guapka [62]
9 less than a number is <em><u>9 - n</u></em>

Where n is that number
6 0
3 years ago
Read 2 more answers
Estimate the cost of 3.2 pounds of apples .The cost of 3.2 pounds of apples is about
Lana71 [14]
138c since 1 pound of apples=43c
7 0
3 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
3 years ago
Please help!! The slope of the line passing through the points (-3, 4) and (4, -1) is ___.
oksano4ka [1.4K]
I hope this helps you


slope=m


(-3,4) x'=-3 y'=4

(4,-1) x"=4 y"=-1


slope formula :


m. (x'-x")=y'-y"


m. (-3-4)=4-(-1)


m.-7=4+1


m=-5/7
4 0
3 years ago
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