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3 years ago

Problem Solving

Mathematics

1 answer:

Misha Larkins [42]3 years ago

5 0

No. Four times five is twenty, so if Andrea has five five dollar bills, she can afford the tennis shoes. <span />

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I need the perimeter and area

kirza4 [7]

Answer:

P: 32

A: 63

Step-by-step explanation:

I hope that thats right..

6 0

3 years ago

P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9

Feliz [49]

Answer:

$x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343$

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > $\frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1$ . Go ahead and simplify.

$x=\frac{9}{d-1};\quad \:d\ne \:1$ .

Now, $\mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}$ .

$P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343$ .

Group the like terms... $1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343$ .

$\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1}$ > $1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343$ .

Now for $1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}$ .

$\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}$ .

Now for $33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}$ .

Thus we then get $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343$ .

Now we want to combine fractions. $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}$ .

$\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2$

$\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}$

$\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}$ .

Expand $81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac$ .

$-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1$ .

$\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}$

Therefore $P=\frac{-297d+378}{\left(d-1\right)^2}+343$ .

Hope this helps!

5 0

3 years ago

HELLPPP PLSSS<br><br>100÷10+3-100+ (6+ 19)+ 3 × -5 -19+100

Ksivusya [100]

Answer:

Step-by-step explanation:

I used a Calculater

3 0

3 years ago

Which are perfect squares? Check all that apply.

den301095 [7]

36, 16, 81, and 64 are all of your perfect squares.

If you need me to explain, just let me know! :)

6 0

3 years ago

Read 2 more answers

Temperature transducers of a certain type are shipped in batches of 50. A sample of 58 batches was selected, and the number of t

andrew11 [14]

Answer:

a) X Freq. Rel Freq.

__________________________

0 7 (7/58) = 0.121

1 10 (10/58)=0.172

2 13 (13/58) =0.224

3 14 (14/58)=0.241

4 6 (6/58)=0.103

5 3 (3/58)=0.052

6 3 (3/58)=0.052

7 1 (1/58) =0.017

8 1 (1/58)=0.017

_________________________

Total 58 1.00

b) $\frac{7+10+13+14+6}{58}= \frac{50}{58}=0.862$

Step-by-step explanation:

Assuming this question: Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined, resulting in the following data:

2,1,2,4,0,1,3,2,0,5,3,3,1,3,2,4,7,0,2,3,

0,4,2,1,3,1,1,3,4,2,3,2,2,8,4,5,1,3,1,

5,0,2,3,2,1,0,6,4,2,1,6,0,3,3,3,6,2,3

(a) Determine frequencies and relative frequencies for the observed values of x = number of nonconforming transducers in a batch. (Round your relative frequencies to three decimal places.)

For this case first we order the dataset on increasing way and we got:

0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3 3 3 3 3 3

4 4 4 4 4 4

5 5 5

6 6 6

And we can construct the following table:

X Freq. Rel Freq.

__________________________

0 7 (7/58) = 0.121

1 10 (10/58)=0.172

2 13 (13/58) =0.224

3 14 (14/58)=0.241

4 6 (6/58)=0.103

5 3 (3/58)=0.052

6 3 (3/58)=0.052

7 1 (1/58) =0.017

8 1 (1/58)=0.017

_________________________

Total 58 1.00

(b) What proportion of batches in the sample have at most four nonconforming transducers? (Round your answer to three decimal places.)

For this case this proportion would be:

$\frac{7+10+13+14+6}{58}= \frac{50}{58}=0.862$

4 0

3 years ago