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3 years ago

What is the average of four tenths and five thousandths?

2 answers:

7 0

$(\frac{4}{10}+\frac{5}{1000}):2 = (\frac{400}{1000}+\frac{5}{1000})*\frac{1}{2} =\\ \\=\frac{405}{1000}*\frac{1}{2}=\frac{405}{2000}=\frac{81}{400}= .2025$

Rama09 [41]3 years ago

5 0

Four tenths is expressed as .4
five thousandths is expressed as .005
the sum of these is .405 and divided by 2 (there are two numbers we added), we get .2025, that is the average

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a line segment has endpoints A (-8,6) and B (4,6). What is the distance in units between point A and B?

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Answer:

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A firm’s marketing manager believes that total sales for next year will follow the normal distribution, with a mean of $3.2 mill

klasskru [66]

Answer:

The sales level that has only a 3% chance of being exceeded next year is $3.67 million.

Step-by-step explanation:

When the distribution is normal, we use 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.

In this question, we have that:

In millions of dollars,

$\mu = 3.2, \sigma = 0.25$

Determine the sales level that has only a 3% chance of being exceeded next year.

This is the 100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So X when Z = 1.88.

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

$1.88 = \frac{X - 3.2}{0.25}$

$X - 3.2 = 0.25*1.88$

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3 years ago

What is the first step in solving the equation x2 – 16/25 = 0?

Keith_Richards [23]

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

x^2-(16/25)=0

Step by step solution :Step 1 : 16 Simplify —— 25 Equation at the end of step 1 : 16 (x2) - —— = 0 25 Step 2 :Rewriting the whole as an Equivalent Fraction :

2.1 Subtracting a fraction from a whole

Rewrite the whole as a fraction using 25 as the denominator :

x2 x2 • 25 x2 = —— = ——————— 1 25

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

x2 • 25 - (16) 25x2 - 16 —————————————— = ————————— 25 25 Trying to factor as a Difference of Squares :

2.3 Factoring: 25x2 - 16

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =
 A2 - AB + BA - B2 =
 A2 - AB + AB - B2 =
 A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 25 is the square of 5
Check : 16 is the square of 4
Check : x2 is the square of x1 

Factorization is : (5x + 4) • (5x - 4)

Equation at the end of step 2 : (5x + 4) • (5x - 4) ——————————————————— = 0 25 Step 3 :When a fraction equals zero : 3.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

(5x+4)•(5x-4) ————————————— • 25 = 0 • 25 25

Now, on the left hand side, the 25 cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
(5x+4) • (5x-4) = 0

Theory - Roots of a product :

3.2 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

3.3 Solve : 5x+4 = 0

Subtract 4 from both sides of the equation :
 5x = -4
Divide both sides of the equation by 5:
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Solving a Single Variable Equation :

3.4 Solve : 5x-4 = 0

Add 4 to both sides of the equation :
 5x = 4
Divide both sides of the equation by 5:
 x = 4/5 = 0.800

x = 4/5 = 0.800
 x = -4/5 = -0.800

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