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Nutka1998 [239]

2 years ago

14

Sunny earns $12 per hour delivering cakes. She worked for x hours this week. Unfortunately, she was charged $15 for a late deliv

ery on Tuesday.
How much money did Sunny earn this week?
Write your answer as an expression.

2 answers:

mixas84 [53]2 years ago

8 0

$12 times X - 15

Hope this is right,
xXharleyquinn04Xx

V125BC [204]2 years ago

3 0

Answer:

12x -15 or 12(x) - 15

Step-by-step explanation:

It's 12x -15 because, 12 is the amount of money sunny makes per hour and X is the amount hours she worked in total. Then, if we subtract the $15 late delivery fee, we will end up with an expression like this, 12x -15 or 12(x) - 15. We can't wright it as an equation because there isn't enough information to do so.

I hope this helps,

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311, 305,299,...Find the 32nd term.

adell [148]

Answer:

503

Step-by-step explanation:

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1 year ago

Ice cream usually comes in 1.5-quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some v

Troyanec [42]

Answer:

(a) The expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.

(b) The expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.

(c) Because the variance of each variable is dependent on the other.

Step-by-step explanation:

The random variable <em>X</em> and <em>Y</em> are defined as follows:

<em>X</em> = amount of ice cream in the box

<em>Y</em> = amount of ice cream scooped out

The information provided is:

E (X) = 48

SD (X) = 1

V (X) = 1

E (Y) = 2

SD (Y) = 0.25

V (Y) = 0.0625

(a)

The total amount of ice-cream served at the party can be expressed as:

<em>X</em> + 3<em>Y</em>.

Compute the expected value of (<em>X</em> + 3<em>Y</em>) as follows:

$E(X+3Y)=E(X)+3E(Y)\\= 48+(3\times2)\\=48+6\\=54$

Compute the variance of (<em>X</em> + 3<em>Y</em>) as follows:

$V(X+3Y) = V (X)+3^{2}V(Y)+2\times 3Cov (X,Y)\\=1+(9\times0.0625)+0\\=1.5625$

Then the standard deviation of (<em>X</em> + 3<em>Y</em>) is:

$SD(X + 3Y) =\sqrt{V(X + 3Y)}\\\sqrt{1.5625}\\=1.25$

Thus, the expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.

(b)

The amount of ice-cream left in the box after scooping out one scoop is represented as follows:

<em>X</em> - <em>Y</em>.

Compute the expected value of (<em>X</em> - <em>Y</em>) as follows:

$E(X-Y)=E(X)-E(Y)\\=48-2\\=46$

Compute the variance of (<em>X</em> - <em>Y</em>) as follows:

$V(X - Y) =V(X)+V(Y)-2Cov(X,Y)\\=1+0.0625-0\\=1.0625$

Then the standard deviation of (<em>X</em> - <em>Y</em>) is:

$SD(X-Y) =\sqrt{V(X -Y)}\\\sqrt{1.0625}\\=1.031$

Thus, the expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.

(c)

The variance of the sum or difference of two variables is computed by adding the individual variances. This is because the variance of each variable is dependent on the others.

7 0

2 years ago

The regression equation you found for the

Viefleur [7K]

Answer:

- the value of 3.915 is the initial number of water lilies

- it is approximately 4, which matches the data

- the value 1.106 is the growth rate

- the rate represents growth for each day

- the percentage growth each day is 10.6%

5 0

1 year ago

Read 2 more answers

En una fábrica de automóviles que trabaja las 24 horas se arman diariamente 24

saul85 [17]

Answer:

a. La ganancia es de $ 4,060,000.00

b. 31 vehículos

Step-by-step explanation:

(a) Los parámetros dados son;

El número de automóviles tipo sedán fabricados = 24

El número de camiones tipo SUV fabricados = 16

El número de camiones tipo VAN fabricados = 12

El número de camionetas pick-up fabricadas = 8

El número de autos deportivos fabricados = 2

La ganancia por la venta de autos tipo sedán = $ 185,000 - $ 140,000 = $ 40,000

La ganancia por la venta de camionetas tipo SUV = $ 320,000 - $ 250,000 = $ 70,000

La ganancia por la venta de camiones tipo VAN = $ 400,000 - $ 310,000 = $ 90,000

La ganancia por la venta de las camionetas pick-up = $ 285,000 - $ 210,000 = $ 75,000

La ganancia por la venta de los autos deportivos = $ 550,000 - $ 400,000 = $ 150,000

La ganancia = 24 * $ 40 000 + 16 * $ 70 000 + 12 * $ 90 000 + 8 * $ 75 000 + 2 * $ 150 000 = $ 4060 000

(b) Por lo que hay una tasa de producción constante, solo la mitad de los automóviles se producirán dentro del período de 12 horas

Por lo tanto, tu fabricado

12 autos sedán, 8 camionetas tipo SUV, 6 camionetas tipo VAN, 4 camionetas pick-up y 1 auto deportivo para hacer un total de 31 vehículos.

6 0

2 years ago

The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,

Gre4nikov [31]

Answer:

A.the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. β = 0.0122

C. β = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

$\mathtt{H_o: \mu = 100}$

$\mathtt{H_1: \mu \neq 100}$

A. If the acceptance region is defined as $98.5 < \overline x > 101.5$ , find the type I error probability $\alpha$ .

Assuming the critical region lies within $\overline x < 98.5$ or $\overline x > 101.5$ , for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is $\mu = 100$

∴

$\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}$

$\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}$

when $\mu = 100$

$\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }$

$\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }$

$\mathtt{\alpha = P ( Z 2.25) }$

$\mathtt{\alpha = P ( Z$

From the standard normal distribution tables

$\mathtt{\alpha = 0.0122+( 1- 0.9878) })$

$\mathtt{\alpha = 0.0122+( 0.0122) })$

$\mathbf{\alpha = 0.0244 }$

Thus, the type 1 error probability is $\mathbf{\alpha = 0.0244 }$

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis $\mathtt{H_o}$

Thus;

β = P( type II error) - P( fail to reject $\mathtt{H_o}$ )

∴

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 103$

$\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }$

$\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}$

From standard normal distribution table

β = 0.0122 - 0.0000

β = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

$\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }$

Given that $\mu = 105$

$\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }$

$\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }$

$\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }$

$\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}$

From standard normal distribution table

β = 0.0000 - 0.0000

β = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0

2 years ago