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Salsk061 [2.6K]
3 years ago
6

5 times the sum of s and 7

Mathematics
1 answer:
EastWind [94]3 years ago
6 0

Answer:

5(s+7) or (s+7)*5

Step-by-step explanation:

Read the question carefully when trying to do these problems. Whenever it states "sum" then you should automatically put the terms behind it (s and 7) together to add them together. This would look like s+7.

Next it says 5 TIMES the sum of the terms therefore you should put s+7 together in parentheses and multiply them by 5 so that you first have the sum of s and 7 in parentheses. This would then look like 5(s+7) or (s+7)*5.

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V=u+at u=2 a=15 t=1/5 what is the value of v
alukav5142 [94]

Answer and explanation:

V originally equals u + at, but when you plug in u, a, and t you get a number answer.

u = 2

a = 15

t = 1/5

Rewrite the expression with the numbers plugged in.

V = (2) + (15)(1/5)  <em>Multiply 15 by 1/5</em>

V = 2 + 3  <em>Add 2 and 3 together</em>

V = 5

<h2><u>So V would equal 5</u></h2>
7 0
3 years ago
Multiply the following polynomials, then place the answer in the proper location on the grid. Write the answer in descending pow
marshall27 [118]
The easiest method of multiplying polynomials, in my experience, is the FOIL method. Honestly, I couldn't tell you what the letters stand for anymore, but I definitely remember how to do it. It's just distribution. 
(6x+1)(5x+8)
Multiply the 6x by 5x and 8, then the 1 by 5x and 8:
30x²+48x+5x+8
Then combine like terms:
30x²+53x+8
5 0
3 years ago
Standardization of Normal Distribution is the process of
n200080 [17]

Answer:

Step-by-step explanation:

Standardizing a normal distribution is to convert a normal distribution to the standard normal distribution. In real-world applications, a continuous random variable may have a normal distribution with a value of the mean that is different from 0 and a value of the standard deviation that is different from 1.

7 0
2 years ago
Evaluate |x + y|, for x = 8 and y = -15.<br><br> 7<br> -7<br> 23<br> -23
natka813 [3]

Answer:

7

Step-by-step explanation:

When evaluating absolute values, a great way to remember what they do is simple. No matter the number, whether it is negative or positive, the number(s) IN THE absolute values are ALWAYS positive.

So, |x + y| is then substituted with the numbers given for the variables.

|x + y|

|8 - 15|

|-7|

7

Hope this helps!

3 0
2 years ago
In this problem we will be dealing with the probability density function (pdf) associated with a continuous random variable, X.
konstantin123 [22]

Answer:

b)\ 0.568

Step-by-step explanation:

Given

f(x) = 6x(1- x);\ 0 \le x \le 1

Required

P(0.3 < x < 0.7)

From the question, we have:

P(a < x < b) = \int\limits^b_a {f(x)} \, dx

So, we have:

P(0.3 < x < 0.7) = \int\limits^{0.7}_{0.3} {6x(1 - x)} \, dx

Open bracket

P(0.3 < x < 0.7) = \int\limits^{0.7}_{0.3} {6x - 6x^2} \, dx

Integrate

P(0.3 < x < 0.7) =  \frac{6x^2}{2} - \frac{6x^3}{3}}|\limits^{0.7}_{0.3}

P(0.3 < x < 0.7) =  3x^2 - 2x^3|\limits^{0.7}_{0.3}

Substitute 0.7 and 0.3 for x

P(0.3 < x < 0.7) =  (3*0.7^2 - 2*0.7^3) - (3*0.3^2 - 2*0.3^3)

Using a calculator, we have:

P(0.3 < x < 0.7) =  (0.784) - (0.216)

Remove brackets

P(0.3 < x < 0.7) =  0.784 - 0.216

P(0.3 < x < 0.7) =  0.568

3 0
2 years ago
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