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Liula [17]
2 years ago
10

Please help I’ll give brainly

Mathematics
1 answer:
Zepler [3.9K]2 years ago
5 0

Answer:

a.100{99.30487yen/1 USD}

Step-by-step explanation:

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1. Which value is the closest to 0.9.997?
Vesna [10]

I think the answer would be D. 10 is closest to 0.9997

4 0
2 years ago
Denis, Vera, and Yash are rock climbers. Yash is connected to Vera by a
grin007 [14]

Answer:

33.14\°

Step-by-step explanation:

Let

Y ----> field of vision that Yash's camera would need

we know that

Applying the law of sines

\frac{sin(Y)}{25}=\frac{sin(41\°)}{30}

Solve for sin(Y)

sin(Y)=\frac{sin(41\°)}{30}(25)

Y=sin^{-1}[\frac{sin(41\°)}{30}(25)]

Y=33.14\°

8 0
3 years ago
Read 2 more answers
Explain the circumstances for which the interquartile range is the preferred measure of dispersion. What is an advantage that th
KatRina [158]

Answer:

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile​ range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

Step-by-step explanation:

Previous concepts

The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

IQR= Q_3 -Q_1

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

Solution to the problem

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile​ range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

8 0
3 years ago
Express the area of the triangle shown below in terms of b and θ only.
Paul [167]
Given a right triangle with base, b and height, h with the angle opposide side h as θ.

The area of a triangle is given by
A= \frac{1}{2} bh

But,
\tan\theta= \frac{h}{b}  \\  \\ \Rightarrow h=b\tan\theta

Therefore, the area of the triangle in terms of <span>b and θ only is given by
A= \frac{1}{2} b(b\tan\theta)= \frac{1}{2} b^2\tan\theta</span>
8 0
3 years ago
Solve for x<br><br> x-4<br> — = 5<br> y+2
yulyashka [42]

I think the question is wrong!

It must be

x-4

___= 5

x+2

Answer: x - 4= 5x + 10

or,-10 -4 = 5x - x

or, -14=4x

or,-14/4= x

Therefore, x = -14/4

5 0
3 years ago
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