Find x -x = 7x

(5 4 – 2) × (–2) = ? a. 22 b. –14 c. 14 d. –22

tamaranim1 [39]

The correct answer is b. -14
:)

6 0

3 years ago

The summer monsoon brings 80% of India's rainfall and is essential for the country's agriculture.

Natasha_Volkova [10]

Answer:

Step 1. Between 688 and 1016mm. Step 2. Less than 688mm.

Step-by-step explanation:

The 68-95-99.7 rule roughly states that in a normal distribution 68%, 95% and 99.7% of the values lie within one, two and three standard deviation(s) around the mean. The z-scores represent values from the mean in a standard normal distribution, and they are transformed values from which we can obtain any probability for any normal distribution. This transformation is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ (1)

$\\ \mu\;is\;the\;population\;mean$

$\\ \sigma\;is\;the\;population\;standard\;deviation$

And x is any value which can be transformed to a z-value.

Then, z = 1 and z = -1 represent values for one standard deviation above and below the mean, respectively; values of z = 2 and z =-2, represent values for two standard deviations above and below the mean, respectively and so on.

Because of the 68-95-99.7 rule, we know that approximately 95% of the values for a normal distribution lie between z = -2 and z = 2, that is, two standard deviations below and above the mean as remarked before.

<h3>Step 1: Between what values do the monsoon rains fall in 95% of all years?</h3>

Having all this information above and using equation (1):

$\\ z = \frac{x - \mu}{\sigma}$

For z = -2:

$\\ -2 = \frac{x - 852}{82}$

$\\ -2*82 + 852 = x$

$\\ x_{below} = 688mm$

For z = 2:

$\\ 2 = \frac{x - 852}{82}$

$\\ 2*82 = x - 852$

$\\ 2*82 + 852 = x$

$\\ x_{above} = 1016mm$

Thus, the values for the monsoon rains fall between 688mm and 1016mm for approximately 95% of all years.

<h3>Step 2: How small are the monsoon rains in the driest 2.5% of all years?</h3>

The driest of all years means those with small monsoon rains compare to those with high values for precipitations. The smallest values are below the mean and at the left part of the normal distribution.

As you can see, in the previous question we found that about 95% of the values are between 688mm and 1016mm. The rest of the values represent 5% of the total area of the normal distribution. But, since the normal distribution is symmetrical, one half of the 5% (2.5%) of the remaining values are below the mean, and the other half of the 5% (2.5%) of the remaining values are above the mean. Those represent the smallest 2.5% and the greatest 2.5% values for the normally distributed data corresponding to the monsoon rains.

As a consequence, the value x for the smallest 2.5% of the data is precisely the same at z = -2 (a distance of two standard deviations from the mean), since the symmetry of the normal distribution permits that from the remaining 5%, half of them lie below the mean and the other half above the mean (as we explained in the previous paragraph). We already know that this value is x = 688mm and the smallest monsoons rains of all year are less than this value of x = 688mm, representing the smallest 2.5% of values of the normally distributed data.

The graph below shows these values. The shaded area are 95% of the values, and below 688mm lie the 2.5% of the smallest values.

3 0

3 years ago

If

arlik [135]

Given:

In a right angle triangle θ is an acute angle and $\tan\theta =\dfrac{3}{5}$ .

To find:

The value of $\cos \theta$ .

Solution:

In a right angle triangle,

$\tan \theta=\dfrac{Perpendicular}{Base}$

We have,

$\tan\theta =\dfrac{3}{5}$

It means the ratio of perpendicular to base is 3:5. Let 3x be the perpendicular and 5x be the base.

By using Pythagoras theorem,

$Hypotenuse=\sqrt{Perpendicular^2+base^2}$

$Hypotenuse=\sqrt{(3x)^2+(5x)^2}$

$Hypotenuse=\sqrt{9x^2+25x^2}$

$Hypotenuse=\sqrt{34x^2}$

$Hypotenuse=x\sqrt{34}$

In a right angle triangle,

$\cos \theta=\dfrac{Base}{Hypotenuse}$

$\cos \theta=\dfrac{5x}{x\sqrt{34}}$

$\cos \theta=\dfrac{5}{\sqrt{34}}$

Therefore, the value of $\cos \theta$ is $\dfrac{5}{\sqrt{34}}$ .

8 0

3 years ago

Need help with this work sheet

Lesechka [4]

Answer:

its alot to explain but i will try to make it as simple as possible

Step-by-step explanation:

your first goal is to make each problem into the form ax^2+bx+c=0

number 1, 2, 7 and 8 is already done for you

now all you have to do is plug in each number in the standard form into the quadtratic formula.

basically at this point you can just use your calculator to do the rest of the work. dont forget parentheses so it doesnt get confused...

or you can perform the algebraic work.. its all just a matter of plugging in the right numbers into the quadratic formula...

cant really do the work for you since im on my phone. but yeah all you need to do step one is transform each problem into ax^2+bx+c=0 form

then step 2, plug in each number in to the quadtratic formula. from there calculate using basic algebraic rules

5 0

3 years ago

A school teacher wants to survey students at her elementary school. She surveys 20 students from each grade level. What type of

Elan Coil [88]

Answer:

Stratified Random sampling

Step-by-step explanation:

When a random observations are selected from a number of individual groups in a particular population, the type of sampling technique is called Stratified Random sampling. Stratified Random sampling begins with the partitioning or splitting or a population into subgroups. A number of random selection are then made from each of the subgroups to form a collection of larger samples. This is different from the simple random sampling technique which makes random selection directly from a larger sample or population without prior partitioning of the population. The different grades of students represents the individual stratum from which random selections are made.

4 0

3 years ago

Find x -x = 7x - 56