ASAP SOMEONE ANSWER THIS ILL GIVE YOU 20 POINTS PLZ!

You purchase pair of pants for $34.60 and a shirt for $12.30. If you need to pay 8% sales tax, what is the final price?

11111nata11111 [884]

Answer:

50.65

Step-by-step explanation:

34.60+12.30=46.90

8%=.08

46.90*.08=3.752

add tax to the total-

46.90+ 3.75= 50.65

4 0

3 years ago

A rectangle has a length of 3x+5 units and a width of x+5 units. A second rectangle has an area of 4x^2-5x+6 square units. What

natima [27]

Answer:

7x^2+15x+31

Step-by-step explanation:

(3x+5)(x+5)

FOIL

3x^2+15x+5x+25

(3x^2+20x+25)+(4x^2-5x+6) = 7x^2+15x+31

7 0

3 years ago

Read 2 more answers

Suppose that the times required for a cable company to fix cable problems in its customers’ homes are uniformly distributed betw

vampirchik [111]

Answer:

a) 60% probability that a randomly selected cable repair visit will take at least 50 minutes

b) 60% probability that a randomly selected cable repair visit will take at most 55 minutes.

c) The expected length of the repair visit is 52.5 minutes.

d) The standard deviation is 7.22 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

For this problem, we have that:

Uniformly distributed between 40 and 65 minutes, so $a = 40, b = 65$

(a) What is the probability that a randomly selected cable repair visit will take at least 50 minutes?

Either it takes less than 50 minutes, or it takes at least 50 minutes. The sum of the probabilities of these events is decimal 1. So

$P(X < 50) + P(X \geq 50) = 1$

The intervals can be open or closed, so

$P(X < 50) = P(X \leq 50)$

We have that

$P(X \geq 50) = 1 - P(X < 50)$

$P(X < x) = \frac{50 - 40}{65 - 40} = 0.4$

$P(X \geq 50) = 1 - P(X < 50) = 1 - 0.4 = 0.6$

60% probability that a randomly selected cable repair visit will take at least 50 minutes

(b) What is the probability that a randomly selected cable repair visit will take at most 55 minutes?

This is $P(X \leq 55)$

$P(X \leq 55) = \frac{55 - 40}{65 - 40} = 0.6$

60% probability that a randomly selected cable repair visit will take at most 55 minutes.

(c) What is the expected length of the repair visit?

The mean of the uniform distribution is:

$M = \frac{a+b}{2}$

So

$M = \frac{40+65}{2} = 52.5$

The expected length of the repair visit is 52.5 minutes.

(d) What is the standard deviation?

The standard deviation of the uniform distribution is

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

So

$S = \sqrt{\frac{(65-40)^{2}}{12}}$

$S = 7.22$

The standard deviation is 7.22 minutes.

5 0

3 years ago

If Theta is an angle in standard position and its terminal side passes through the point

Alex777 [14]

Answer: - 9/8

Step-by-step explanation:

6 0

3 years ago

4. Tests are performed to determine the Tool Life of a high speed steel single edge cutting tool. At a cutting speeds (v) of 180

podryga [215]

Answer:(a) n=0.233

Step-by-step explanation:

Given

We Know according to Taylor tool life

$VT^n=c$

V=cutting speed

T=tool life

n=tool life exponent

c=constant

$180\times \left ( 4.25\right )^n=110\times \left ( 35\right )^n$

$\frac{180}{110}=\left ( \frac{35}{4.25}\right )^n$

$1.636=\left ( 8.235\right )^n$

taking log both side

$\ln \left ( 1.636\right )=n\ln \left ( 8.235\right )$

n=0.233

3 0

3 years ago