Ask question

Ask question

All categories

3 years ago

7

Cody Summer sells tires for tractor trailers. He earns 4.5 percent commission on the first $6,000, 8 percent on the next $6,000,

and 12 percent on sales over $12,000. Last month he sold $18,000 worth of tires. What was his total commission?

1 answer:

Paladinen [302]3 years ago

7 0

The Answer is $1,470.

You might be interested in

Simplify (x^6)^2 • x^3

givi [52]

Answer:

x^15

Step-by-step explanation:

Recall these rules of exponents:

(a^m)^n = a^mn

a^m • a^n = a^(m + n)

(x^6)² • x³ = x^(2 • 6) • x³ = x^12 • x³ = x^(12 + 3) = x^15

8 0

2 years ago

On one day a pet store sells 2 birds 6 gerbils 3 fish and 3 hamsters whats the data and relative frequency as a percent?

bonufazy [111]

Answer:

Birds $F_b=14.3\%$

Gerbils $F_g=42.9\%$

Fish $F_f=21.4\%$

Hamsters $F_h=21.4\%$

Step-by-step explanation:

From the question we are told that:

Sales

Number of birds $n_b=2$

Number of gerbils $n_g=6$

Number of fish $n_f=3$

Number of hamsters $n_h=3$

Generally the frequency for birds $F_b$ is mathematically given b

$F_b=\frac{n_b}{n}*100$

$F_b=\frac{2}{14}*100$

$F_b=14.3\%$

Generally the frequency for Gerbils $F_g$ is mathematically given b

$F_g=\frac{n_g}{n}*100$

$F_g=\frac{6}{14}*100$

$F_g=42.9\%$

Generally the frequency for fish $F_f$ is mathematically given b

$F_f=\frac{n_f}{n}*100$

$F_f=\frac{3}{14}*100$

$F_f=21.4\%$

Generally the frequency for fish $F_h$ is mathematically given b

$F_h=\frac{n_h}{n}*100$

$F_h=\frac{3}{14}*100$

$F_h=21.4\%$

5 0

3 years ago

Anthony has been saving nickels dimes quarters over the last year he knows that he has twice as many nickels as quarters. change

OLga [1]

Answer:

The number of nickel's coin is 150

The number of dime's coin is 35

The number of quarter's coin is 75

Step-by-step explanation:

Let

x ----> the number of nickel's coin

y ---> the number of dime's coin

z ---> the number of quarter's coin

Remember that

$1\ nickel=\$0.05$

$1\ dime=\$0.10$

$1\ quarter=\$0.25$

we know that

$x+y+z=260$ ----> equation A

$0.05x+0.10y+0.25z=29.75$ ----> equation B

$x=2z$ ----> equation C

substitute equation C in equation A and equation B

$2z+y+z=260$

$3z+y=260$ ---> equation A1

$0.05(2z)+0.10y+0.25z=29.75$

$0.35z+0.10y=29.75$ -----> equation B1

Solve the system of equations A1 and B1 by graphing

Remember that the solution is the intersection point both graphs

The solution is the point (75,35)

see the attached figure

so

z=75, y=35

<em>Find the value of x</em>

$x=2(75)=150$

therefore

The number of nickel's coin is 150

The number of dime's coin is 35

The number of quarter's coin is 75

8 0

3 years ago

Lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. a bank conducts inter

Otrada [13]

Part A:

Given that lie <span>detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector correctly determined that a selected person is saying the truth has a probability of 0.85
Thus p = 0.85

Thus, the probability that </span>the lie detector will conclude that all 15 are telling the truth if <span>all 15 applicants tell the truth is given by:

</span> $P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(15)={ ^{15}C_{15}(0.85)^{15}(0.15)^0} \\ \\ =1\times0.0874\times1=0.0874$
<span>

</span>Part B:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
Thus p = 0.15

Thus, the probability that the lie detector will conclude that at least 1 is lying if all 15 applicants tell the truth is given by:

$P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(X\geq1)=1-P(0) \\ \\ =1-{ ^{15}C_0(0.15)^0(0.85)^{15}} \\ \\ =1-1\times1\times0.0874=1-0.0874 \\ \\ =0.9126$

Part C:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The mean is given by:

$\mu=npq \\ \\ =15\times0.15\times0.85 \\ \\ =1.9125$

Part D:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The <span>probability that the number of truthful applicants classified as liars is greater than the mean is given by:

</span> $P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\ \\ 1-[P(0)+P(1)]$
<span>
</span> $P(1)={ ^{15}C_1(0.15)^1(0.85)^{14}} \\ \\ =15\times0.15\times0.1028=0.2312$ <span>
</span>

8 0

3 years ago

PLEASE HELLPPPPPPPPPPPPP!!!!!!!!

Slav-nsk [51]

The first answer is 25x + 45 = c
The second one is y=3x
Hopes this helps

6 0

3 years ago