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nadezda [96]
3 years ago
12

Use the Distributive Property to rewrite 5(15 + x).

Mathematics
2 answers:
aivan3 [116]3 years ago
7 0

Answer:

75+5x

Step-by-step explanation:

Distribute the 5, or multiply everything in the parentheses by 5

5(15+x)

5*15 +5*x

75+ 5x

liq [111]3 years ago
5 0

Answer:

The answer is -15x-10

Step-by-step explanation:

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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 <em>68-95-99.7 rule </em>roughly states that in a <em>normal distribution</em> 68%, 95% and 99.7% of the values lie within one, two and three standard deviation(s) around the mean. The z-scores <em>represent values from the mean</em> in a <em>standard normal distribution</em>, 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 <em>x</em> is any value which can be transformed to a z-value.

Then, z = 1 and z = -1 represent values for <em>one standard deviation</em> 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 <em>driest of all years</em> 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 <em>symmetrical</em>, 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 <em>x </em>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 <em>x</em> = 688mm and the smallest monsoons rains of all year are <em>less than this value of x = </em><em>688mm</em>, 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
A container contains 12 diesel engines. The company chooses 5 engines at​ random, and will not ship the container if any of the
Rufina [12.5K]

Answer:\frac{^{10}C_5}{^{12}C_5}

Step-by-step explanation:

Given a total of 12 diesel engines

out of which 2 are defective

Company have to choose 5 good engines to ship.

Therefore no of ways in good engines are selected is ^{10}C_5

no of ways in which any 5 engines are selected from a total of 12 engines is

^{12}C_5

Therefore the required probability =\frac{^{10}C_5}{^{12}C_5}

=\frac{7}{22}=0.318

3 0
3 years ago
There are 26 students in Jenna's class. Jenna brought in 16 cupcakes. Which equation can be used to find the number of cupcakes
Olin [163]

Answer:

Step-by-step explanation:

16+n=26

26-16=n

4 0
3 years ago
Read 2 more answers
Find the least common denominator (lcd) of 1/3 and 2/9 .
OLEGan [10]
The lowest common denominator of 1/3 and 2/9 = 9
8 0
3 years ago
Don't steal points
Tomtit [17]

Answer:

The ceiling is lower than 3 meters.

Step-by-step explanation:

In the above, the product is <em>less than</em> 3. In this case, it will not include 3 as a choice, as if it would, the \leq would be used.

This means that x would be any number that is less than 3. The only choice that gives a choice that means less than by the terms definition is A) The ceiling is lower than 3 meters.

~

6 0
3 years ago
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