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ASHA 777 [7]
2 years ago
9

Hilda was simplifying some numerical expressions and made each of the following sequences of calculations. Name the mathematical

property, operation, or idea that justifies how Hilda went from each step to the next step, some of the steps are done for you. Look at the Math Notes found on page 190 for definitions. Problem
Problem 1: 5 · (−4/3) · (2/5)
= (−4/3) · 5 · (2/5)
= (−4/3) · (5 · (2/5))
= (−4/3) · (2/1) = −8/3 = −2*2/3
Problem 2: 17 + 29 + 3+ 1
= 17 + 3 + 29 + 1
= (17 + 3) + (29 + 1)
= 20 + 30 Added groups
= 50 Added terms
Mathematics
2 answers:
Galina-37 [17]2 years ago
4 0

Answer:

Problem 1:

5 · (−4/3) · (2/5)

= (−4/3) · 5 · (2/5)   <em>Conmutative Property of Multiplication</em>

= (−4/3) · (5 · (2/5))  <em>Associative Property of Multiplication</em>

= (−4/3) · (2/1) = −8/3 = −2*2/3<em> Multiplied fractions and extracted common factor</em>

Problem 2:

17 + 29 + 3+ 1

= 17 + 3 + 29 + 1  <em>Conmutative Property of Addition</em>

= (17 + 3) + (29 + 1)  <em>Associative Property of Addition</em>

= 20 + 30 Added groups

= 50 Added terms

Step-by-step explanation:

<u>For the Problem 1:</u>

In the first step, Hilda applied the <em>Conmutative Property of Multiplication</em>, because she changed the order of the numbers in the product

In the second step, she applied the <em>Associative Property of Multiplication, </em>because she agrouped the product of 5 and<em> </em>2/5 to perform it sepparately

In the third step, she calculated<em> the product of the fractions</em> -4/3 and 2/1, then she extracted 2 as a <em>common factor</em> to express the fraction as -2*2/3

<u>For the Problem 2:</u>

In the first step, Hilda applied the <em>Conmutative Property of Addition, </em>because she changed the order of the numbers in the sum

In the second step, she applied the<em> Associative Property of Addition, </em>because she associated the addition of 17 and 3 and the addition of 29 and 1, to calculate them in groups.

<em />

erica [24]2 years ago
3 0

Answer:

The reason for each statement is shown below.

Step-by-step explanation:

Commutative Property of Multiplication:

a\times b=b\times a

Associative Property of Multiplication:

a\times (b\times c)=(a\times b)\times c

Commutative Property of Addition:

a+b=b+a

Associative Property of Addition:

a+(b+c)=(a+b)+c

Problem 1:

The given expression is

5\cdot (-\frac{4}{3})\cdot(\frac{2}{5})

It can be written as

[5\cdot (-\frac{4}{3})]\cdot(\frac{2}{5})

[(-\frac{4}{3})\cdot 5]\cdot (\frac{2}{5})          (Commutative Property of Multiplication)

(-\frac{4}{3})\cdot [5\cdot (\frac{2}{5})]        (Associative Property of Multiplication)

(-\frac{4}{3})\cdot \frac{2}{1}=-\frac{8}{3}=-2\frac{2}{3}        (Simplification)

Problem 2:

The given expression is

17 + 29 + 3+ 1

It can written as

17 + (29 + 3)+ 1

17 + (3+29)+ 1         (Commutative Property of Addition)

(17 +3)+(29+ 1)        (Associative Property of Addition)

20+30          (Added groups)

50          (Added terms)

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The batting Wang Xiu Ying uses to fill quilts has a thermal conductivity rate of 0.030.030, point, 03 watts (\text{W})(W)left pa
alexgriva [62]

Answer:

0.0003W/cm°C

Step-by-step explanation:

The question is not properly written. Here is the correct question.

The batting wang xiu ying uses to fill quilts has a thermal conductivity rate of 0.03 watts (W) per meter(m) per degree celsius. what is the batting thermal conductivity when w/cm•c

Given the thermal conductivity in W/m°C to be 0.03W/m°C

We are to rewrite the value in W/cm°C

The difference is the unit. The only thing we need to do is to simply convert the unit (metres) in W/m°C to centimeters (cm)

Since 100cm = 1m, 0.03W/m°C can be expressed as shown below;

= 0.03W/m°C

= 0.03 × W/1m×°C

Note that 1m = 100cm, substituting this conversion into the expression, it will become;

= 0.03 × W/100cm × °C

= 0.03/100 × W/cm°C

= 0.0003W/cm°C

Hence the battling thermal conductivity in W/cm°C is 0.0003W/cm°C

4 0
2 years ago
Sarah is planning to spend a week at her friend's summer house in Miami Beach. All meals will be provided by her friend's parent
Rus_ich [418]

Answer:

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Step-by-step explanation:

5 0
2 years ago
The value of a rare painting has increased each year since it was found at a garage sale. The value of the painting is modeled b
solmaris [256]

799 represents the initial value of the painting

The painting will be worth $926 after 5 years to the nearest dollar

Step-by-step explanation:

The form of the exponential function is f(x)=a(b)^{x} , where

  • a is the initial value ⇒ (at x = 0)
  • b is the growth/decay factor ⇒ (rate of change)
  • If b > 1, then the function is growth ⇒ (increasing)
  • If 0 < b < 1, then the function is decay ⇒ (decreasing)

The value of a rare painting has increased each year since it was

found at a garage sale

The value of the painting is modeled by the function f(x)=799(1.03)^{x}

We need to know what 799 represents and what the painting will

be worth after 5 years

∵ f(x)=799(1.03)^{x}

∵ The form of the function is f(x)=a(b)^{x}

- By comparing the two forms

∴ a = 799 and b = 1.03

∵ a is the initial value at x = 0

∴ 799 represents the initial value of the painting

∵ x represents the number of years

∴ x = 5

- Substitute the value of x in the function by 5

∵ f(5)=799(1.03)^{5}

∴ f(x) = 926.26

∴ The painting will be worth $926 after 5 years to the nearest dollar

799 represents the initial value of the painting

The painting will be worth $926 after 5 years to the nearest dollar

Learn more:

You can learn more about the functions in brainly.com/question/10382470

#LearnwithBrainly

6 0
2 years ago
After years of maintaining a steady population of 32,000, the population of a town begins to grow exponentially. After 1 year an
galben [10]

Answer:

y=32000(1+0.08)^x

Step-by-step explanation:

Exponential growth function is y=a(1+r)^x

Where 'a' is the initial population

r is the rate of growth and x is the time period in years

a steady population of 32,000. So initial population is 32,000

an increase of 8% per year. the rate of increase is 8% that is 0.08

a= 32000 and r= 0.08

Plug in all the values in the general equation

y=a(1+r)^x

y=32000(1+0.08)^x

y=32000(1+0.08)^x

4 0
2 years ago
Read 2 more answers
A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm
AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, P\left(\frac{E}{A}\right)=0.02

Now, from the conditional probability,

P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}

\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)

\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

P\left(\frac{E}{A}\right)=0.40

and \Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)

This is the probability of disregarding the partially defective parts by inspection machine.

P\left(\frac{E}{A}\right)=0.98

and \Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

=P(E\cap C)

= 0.0686 [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)

\Rightarrow P(E)=0.018+0.012+0.0686

\Rightarrow P(E)=0.0986

So, the total probability that the part produced get shipped

=1-P(E)=1-0.0986=0.9014

The probability that the part is good (either defect free or slightly defective)

=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)

=(0.9-0.018)+(0.03-0.012)

=0.9

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}

=\frac{0.9}{0.9014}

=0.9984

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

5 0
2 years ago
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