Is line 1 and 2 parallel perpendicular lines or neither<br> line 1 6x+10y=-4<br> line 2 3y-=-5x+4

In remodeling a house an architect finds that by adding the same amount to each dimension of a 15ft by 19ft rectangular room, th

earnstyle [38]

Answer:

Step-by-step explanation:

new length=15+x

width=19+x

then area=(15+x)×(19+x)=285+15x+19x+x²=x²+34x+285 ft²

original area=15×19=285 ft²

then 285+98=x²+34x+285

or

x²+34x-98=0

x²+34x+17²=98+17²

(x+17)²=98+289=387

x+17=√387=3√43

x=3√43-17 ft

8 0

3 years ago

For a school play, 739 tickets valued at $857 were sold. Some cost $1 and others cost a $1.50. How many $1 tickets were sold?

balu736 [363]

Answer:

503 $1 tickets sold.

Step-by-step explanation:

Use two equations

Let x = number of $1 tickets sold

Let y = number of $1.50 tickets sold

x + y = 739

1x + (1.5)y = 857

First equation ==> y = 739 - x

Plug this into the second equation

x + (1.5)(739 - x) = 857

x + 1108.5 - 1.5x = 857

- 0.5x = -251.5

x = 503

There were 503 $1 tickets sold.

To find the number of $1.50 tickets, just plug this value of x into either one of the equations.

(503) + y = 739 (739 - 503 = 236)

y = 236

There were 236 $1.50 tickets sold.

3 0

4 years ago

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

3 years ago

Charlie runs ata speed of 3 yards per second. about how many miles per hour does charlie run?

DochEvi [55]

(3 yard/sec) x (3,600 sec/hr) x (1 mile / 1,760 yard) =

(3 x 3,600 / 1,760) (mile/hour) = 6.136... mph (rounded)

8 0

4 years ago

The equation of the line of best fit is y=3x+2. x=the number of ice cream sales and y= the number of days. Answer the following

Nina [5.8K]

Answer:

all I know is how to get y but y=122

Step-by-step explanation:

y=3(40)+2

y=120+2

y=122

sorry but I dont know what to do for x

3 0

3 years ago

Is line 1 and 2 parallel perpendicular lines or neither line 1 6x+10y=-4 line 2 3y-=-5x+4