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vfiekz [6]
2 years ago
7

Solution to the inequality 16x−33x<37x+27.

Mathematics
1 answer:
astra-53 [7]2 years ago
3 0

<em><u>The solution to the inequality is:</u></em>

x>\frac{-1}{2}

<em><u>Solution:</u></em>

Given inequality is:

16x-33x

We have to find the solution to given inequality

\mathrm{Add\:similar\:elements:}\:16x-33x=-17x

-17x

\mathrm{Subtract\:}37x\mathrm{\:from\:both\:sides}

-17x-37x

Simplify the above inequality

-54x

\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}

Remember that, change the inequality sign if you divide or multiply both sides by a negative number

If you divide or multiply both sides by a positive number,the inequality sign will not change

\left(-54x\right)\left(-1\right)>27\left(-1\right)\\\\54x>-27

\mathrm{Divide\:both\:sides\:by\:}54

\frac{54x}{54}>\frac{-27}{54}\\\\x>-\frac{1}{2}

Thus the solution to inequality is found

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Tim has an after-school delivery service that he provides for several small retailers in town. He uses his bicycle and charges $
nlexa [21]

Answer:

<em>$4.84</em>

<em></em>

Step-by-step explanation:

Given that

For delivery within 1\frac{1}2 mi, charges = $1.25

For delivery within 1\frac{1}2 mi - 1\frac{3}4 mi, charges = $1.70

For delivery within 1\frac{3}4 mi - 2 mi, charges = $2.15

and

so on.

i.e.

For delivery within 2 mi - 2\frac{1}4 mi, charges = $2.60

For delivery within 2\frac{1}4 mi - 2\frac{1}2 mi, charges = $3.05

For delivery within 2\frac{1}2 mi - 2\frac{3}4 mi, charges = $3.50

For delivery within 2\frac{3}4 mi - 3 mi, charges = $3.95

For delivery within 3 mi - 3\frac{1}4 mi, charges = $4.40

So, every 0.25 mi or \frac{1}4 mi increase in distance, there is an increase of $0.45 in the charges.

It is given that there is increase of 10% in the rates.

i.e. for every 0.25 mi increase in distance, there is an increase of 0.45*1.10 = $0.495 in the charges.

The distance of 3\frac{1}8 miles lies within the range 3 mi - 3\frac{1}4 mi.

So actual charges after increase of 10%

\Rightarrow \$4.40 \times \dfrac{110}{100}\\\Rightarrow \$4.40 \times 1.10\\\Rightarrow \bold{\$4.84}

6 0
2 years ago
Triangle PQR has coordinates P (2, 4), Q (-2, 4), R (0,-6).
alexdok [17]

Answer: P’ (5,10) , Q’ (-5,10) , R’ ( 0,-15)

Step-by-step explanation:

Just multiply 2.5 by the x,y

4 0
1 year ago
Read 2 more answers
Daniel deposited $500 into a savings account and after 8 years, his investment is worth $807.07. The equation A = d(1.005)12t mo
shtirl [24]

Answer:

Step-by-step explanation:

The equation A = d(1.005)^12t modelling the value of Daniel’s investment shows a monthly compounded interest. This means that the interest is compounded 12 times in a year.

We can confirm by inputting the given values

t = 8 years

d = 509

Therefore,

A = 500(1.005)12 × 8

A = 500(1.005)^96

A = $807.07

Therefore, the true statements are

Increases

Exponential

Never Decrease

5 0
2 years ago
Students at a private liberal arts college are classified as being freshmen, sophomores, juniors or seniors, and also according
labwork [276]
The total number of possible classifications for the students of this college is found by multiplying 4 (which is the classification for the year level:freshman, sophomore, juniou, senior) and 2 (which is the number of sexes: female and male). So 4 x 2 = 8. There are eight possible classifications, which are:
(Male, Freshman)
(Male, Sophomore)
(Male, Junior)
(Male, Senior)
(Female, Freshman)
(Female, Sophomore)
(Female, Junior)
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5 0
2 years ago
A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur
11111nata11111 [884]
(a) 0.059582148 probability of exactly 3 defective out of 20

 (b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

  (a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

 0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)

 = 0.05^3 * 0.95^17 * 20! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

 = 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)

 = 0.05^3 * 0.95^17 * 20*19*3

 = 0.000125* 0.418120335 * 1140

 = 0.059582148

  (b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.

 The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)

 = 0.05*0.95^3*24/(1!3!)

 = 0.05*0.857375*24/6

 = 0.171475

 
 So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
3 0
2 years ago
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