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3 years ago

NEED HELP PLZ ASAP WHAT IS X -6(-14+3x)-2x=144

Mathematics

2 answers:

NISA [10]3 years ago

8 0

Answer:

Step-by-step explanation:

x=-3

vlabodo [156]3 years ago

4 0

Answer:

Its -3

Step-by-step explanation:

Hope this helped :)

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Two movie tickets and 3 snacks are $24.two movie tickets and 5 snacks are $28.how much is a movie ticket and how much is a snack

maks197457 [2]

Answer:

S: $0.18

T: $11.73

Step-by-step explanation:

T=tickets

S=snacks

2t+3s=24

5s=28

Divide 5 with 28.

Roughly $0.18

s=0.18.

2t+3(0.18)=24

2t+0.54=24

Subtract 0.54:

2t+0.54-0.54=24-0.54

2t=23.46

Divide 2:

t=11.73

So A snack is $0.18, and a ticket is $11.73

Hope this helped!

4 0

3 years ago

Everything I don’t understand

irina1246 [14]

Answer:

what don't you understand

8 0

3 years ago

Read 2 more answers

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s). Line m has no y

goblinko [34]

If the line does not have a y int, then it means it does not cross the y axis....that means it is a vertical line.....its x int is (3,0)...so this equation for the line is x = 3

if the line does not have an x int, it means it does not cross the x axis...that means it is a horizontal line...its y int is (0,-4)...so the equation for this line is
y = -4

4 0

3 years ago

Read 2 more answers

How heavy a load (pounds) is needed to pull apart pieces of wood 4 inches long and 1.5 inches square? Here are data from student

statuscvo [17]

Answer:

a. (29526.192 , 32155.808)

b. (29673.9301 , 32008.0699)

Step-by-step explanation:

on calculation, we find that sample mean = 30843 and sample standard deviation = 3018.504, and n=20

Sample Mean = 30841

SD = 3000

Sample Size (n) = 20

Standard Error (SE) = SD/root(n) = 670.8204

alpha (a) = 1 - 0.95 = 0.05

z critical value for 95% confidence interval:

z(a/2) = z(0.025) = 1.96

Margin of Error (ME) = z(a/2) x SE = 1314.808

95% confidence interval is given by:

Sample Mean +/- (Margin of Error) = 30841 +/- 1314.808

= (29526.192 , 32155.808)

2)when std dev of population is not known, we use sample's, but we have to use t instead of z

Sample Mean = 30841

SD = 3018.504

Sample Size (n) = 20

Standard Error (SE) = SD/root(n) = 674.958

alpha (a) = 1-0.9 = 0.1

we use t-distribution as population standard deviation is unknown

t(a/2, n-1 ) = 1.7291

Margin of Error (ME) = t(a/2,n-1)x SE = 1167.0699

90% confidence interval is given by:

Sample Mean +/- (Margin of Error)

30841 +/- 1167.0699 = (29673.9301 , 32008.0699)

4 0

3 years ago

A car rental agency has 150 cars. The owner finds that at a price of $48 per day, he can rent all the cars. For each $2 increase

hram777 [196]

Given that for each $2 increase in price, the demand is less and 4 fewer cars are rented.

Let x be the number of $2 increases in price, then the revenue from renting cars is given by
$(48 + 2x) \times (150 - 4x)=7,200+108x-8x^2$ .

Also, given that for each car that is rented, there are routine maintenance costs of $5 per day, then the total cost of renting cars is given by
$5(150-4x)=750-20x$

Profit is given by revenue - cost.
Thus, the profit from renting cars is given by
 $(7,200+108x-8x^2)-(750-20x)=6,450+128x-8x^2$

For maximum profit, the differentiation of the profit function equals zero.
i.e.
 $\frac{d}{dx} (6,450+128x-8x^2)=0 \\ \\ 128-16x=0 \\ \\ x= \frac{128}{16} =8$

The price of renting a car is given by 48 + 2x = 48 + 2(8) = 48 + 16 = 64.

Therefore, the rental charge will maximize profit is $64.

3 0

3 years ago