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4 years ago

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A construction worker bought several bottles of juice for $3 at the comedian store she paid for them the $20 bills if J represen

ts a number of bottles of juice write an expression for the change she would receive

1 answer:

irina [24]4 years ago

4 0

Expression 20 – 3J represents the change in dollars, which construction worker will receive after buying J bottles of juice.

<u>Solution:</u>

Given that

A construction worker bought several bottles of juice for $3 at the comedian store

She paid for them the $20 bills

Number of bottles of juice is represented by variable J

Need to write an expression for the change she receives.

From given information

Price of 1 juice bottle = $3

$\text { So price of J juice bottles }=\mathrm{J} \times \text { Price of } 1 \text { juice bottle }$

$=\mathrm{J} \times 3=3 \mathrm{J}$

<em>Change she receives = Amount she paid - price of J juice bottles </em>

=> Change she receives = 20 – 3J

Hence expression 20 – 3J represents the change in dollars, which construction worker will receive after buying J bottles of juice.

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An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

3 years ago

What is the unit rate $1.99 for 30 oz to nearest tenth

KatRina [158]

Answer:

$0.10/oz

Step-by-step explanation:

Unit rate simply means the amount per 1 unit of something.

In this case, we want to find the amount of money for 1 oz of something.

So, we divide 1.99 by 30 and divide 30 by 30 to get:

(1.99 / 30) / (30 / 30) = $0.0663 / 1 oz

Rounded to the nearest tenth, 0.0663 is about 0.1.

Thus, the answer is $0.10/oz.

6 0

3 years ago

Jeff and Kirk can build a 75-ft retaining wall together in 4 hours. Because Jeff has more experience, he could build the wall

oee [108]

It will take Kirk about 8 hours 32 minutes to build the wall by himself.

<h3>What is Quadratic Polynomial?</h3>

Quadratic equations are second-degree algebraic expressions and are of the form ax² + bx + c = 0.

Here, Let k represent Kirk's time (in hours) to build the wall by himself. Then (k-1) is the time it takes for Jeff to build it. Their working-together rate in "walls per hour" is ...

1/k +1/(k -1) = 1/4

Multiplying by 4k(k-1), we have ...

4(k-1) +4k = k(k -1)

k² -9k = -4 . . . . . . . . subtract 8k and simplify

k² -9k +20.25 = 16.25 . . . . . add (9/2)² to complete the square

k -4.5 = √16.25 . . . . . . . . . take the square root

k = 4.5 +√16.25 ≈ 8.531129 . . . hours

The fractional hour is ...

0.531129 × 60 min ≈ 31.9 min ≈ 32 min

Thus, It will take Kirk about 8 hours 32 minutes to build the wall by himself.

Learn more about Quadratic Polynomial from:

brainly.com/question/2350120

#SPJ1

8 0

2 years ago

Read 2 more answers

In the figure, one side of the rectangle divides the circle in half.

Setler79 [48]

A(rectangle)=5*12=60
A(circle)=(pi)(r^2)
r=(1/2)(5)=2.5
A(circle)=(3.14)(2.5)^2
A(circle)=19.625
Shaded Circle=1/2
(1/2)(19.625)=9.8125
A of unshaded=60-9.8125
A=50.1875 ft

6 0

4 years ago

Scores on a test are normally distributed with a mean of 81.2 and a standard deviation of 3.6. What is the probability of a rand

Misha Larkins [42]

<u>Answer:</u>

The probability of a randomly selected student scoring in between 77.6 and 88.4 is 0.8185.

<u>Solution:</u>

Given, Scores on a test are normally distributed with a mean of 81.2

And a standard deviation of 3.6.

We have to find What is the probability of a randomly selected student scoring between 77.6 and 88.4?

For that we are going to subtract probability of getting more than 88.4 from probability of getting more than 77.6

Now probability of getting more than 88.4 = 1 - area of z – score of 88.4

$\mathrm{Now}, \mathrm{z}-\mathrm{score}=\frac{88.4-\mathrm{mean}}{\text {standard deviation}}=\frac{88.4-81.2}{3.6}=\frac{7.2}{3.6}=2$

So, probability of getting more than 88.4 = 1 – area of z- score(2)

= 1 – 0.9772 [using z table values]

= 0.0228.

Now probability of getting more than 77.6 = 1 - area of z – score of 77.6

$\mathrm{Now}, \mathrm{z}-\text { score }=\frac{77.6-\text { mean }}{\text { standard deviation }}=\frac{77.6-81.2}{3.6}=\frac{-3.6}{3.6}=-1$

So, probability of getting more than 77.6 = 1 – area of z- score(-1)

= 1 – 0.1587 [Using z table values]

= 0.8413

Now, probability of getting in between 77.6 and 88.4 = 0.8413 – 0.0228 = 0.8185

Hence, the probability of a randomly selected student getting in between 77.6 and 88.4 is 0.8185.

4 0

3 years ago