What's the value of x in the figure?

4688-3) -30X-5 CX-5)

pshichka [43]

Answer:

x = 13/3

Step-by-step explanation:

4(8x-3) - 30x = 5(x-5)

Distribute

32x - 12 - 30x = 5x -25

Combine like terms

2x -12 = 5x-25

Subtract 2x from each side

2x-12-2x = 5x-2x -25

-12 = 3x-25

Add 25 to each side

-12+25 = 3x-25+25

13 = 3x

Divide each side by 3

13/3 = x

4 0

3 years ago

Read 2 more answers

please help me.... A painting company charges $250 base plus $16 per hour. Another painting company charges $210 base plus $18 p

dem82 [27]

Basically its 250 + 16x = 210 + 18x where x = the number of hours doing the job.

Solving, we get 40 = 2x, x =20

The jobs has to be 20 hours long in order for their two painting jobs to be equally priced.

Hope this helped!

3 0

2 years ago

the ratio of pennies to quarters in a piggy bank is 14:3. if there are 51 quarters in the bank, how many pennies are there

malfutka [58]

The number of pennies in bank is 42

Solution:

Given that ratio of pennies to quarters in a piggy bank is 14 : 3

Let 14x be the number of pennies in piggy bank

Let 3x be the number of quarters in piggy bank

To find: number of pennies

Given that there are 51 quarters

number of pennies in piggy bank + number of quarters in piggy bank = 51

14x + 3x = 51

17x = 51

x = 3

number of pennies = 14x = 14(3) = 42

Thus the number of pennies in bank is 42

4 0

2 years ago

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Houston. The stud

iris [78.8K]

Answer:

$-0.048$

Do not reject $H_0:P_1-P_2=0$

Step-by-step explanation:

From the question we are told that

Sample size $n_1=350$

Sample size $n_2=250$

Sample proportion 1 $\hat P= \frac{25}{350} =>0.07$

Sample proportion 2 $\hat P= \frac{19}{250} =>0.076$

95% confidence interval

Generally for 95% confidence level

Level of significance

$\alpha = 1-0.95=>0.05$

$\alpha /2=\frac{0.05}{2} =>0.025$

Therefore

$Z_a_/_2=1.96$

Generally the equation for confidence interval between $P_1 - P_2$ is mathematically given as

$(\hat P_1-\hat P_2)\pm Z_a_/_2\sqrt{\frac{\hat P_1(1-\hat P_1)}{n_1}+\frac{\hat P_2(1-\hat P_2)}{n_2} }$

$(0.07-0.076)\pm 1.96\sqrt{\frac{0.07(1-0.07)}{350}+\frac{0.076(1-0.076)}{250} }$

$(0.07-0.076)\pm 1.96\sqrt{4.66896*10^-^4 }$

$(-0.006)\pm 0.042$

$(-0.006)- 0.042=>-0.048$

$(-0.006)+ 0.042=>0.036$

Therefore

Confidence interval is

$-0.048$

Conclusion

Given the confidence interval has zero

Therefore do not reject $H_0:P_1-P_2=0$

6 0

2 years ago

Consider the functions f(x)=(ax^2-b)/(x^2-4) and g(x)=(x^2-cx+d). if the limit as x approaches 3 equals 6, what is the relations

Dmitry_Shevchenko [17]

$6=\dfrac{a\cdot3^2-b}{3^2-4}\\\\
6=\dfrac{9a-b}{5}\\\\\
9a-b=30\\\\
\boxed{b=9a-30}$

6 0

2 years ago