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3 years ago

Carly paid $17.50 for 7 gallons of gas and Jade paid $45 for 15 gallons of gas. Which of the following statements is true?

1 answer:

6 0

First we have to find how much they spent on each gallon.
To find out how much Carly paid for each gallon we have to divide her total price by haw many gallons she got.
$17.50/7=$2.50 per gallon
Likewise we do the same for Jade
$45/15=$3

So the answer will be Jade paid more per gallon than Carly

You might be interested in

Using the distributive property to factorize the equation 3x^2+24x=0, you get

Natalka [10]

Answer:

Step-by-step explanation:

$3x^2+24x=0\\factor: 3x(x+8)=0\\x=0, -8$

6 0

3 years ago

Read 2 more answers

Area=<br> Help me please!! Thanks :)

Savatey [412]

Answer:

80π

Step-by-step explanation:

Circle O with radius IO:

large radius = IO = 12 = R

Circle O with diameter JK

IJ = JK = KL = 2 × IO = 24

IJ = JK = KL = 24/3 = 8

small radius = 8/2 = 4 = r

The shaded area is a semicircle of radius 12 minus a semicircle of radius 4 plus two semicircles of radius 4.

That is the same as

1 semicircle of radius 12 plus 1 semicircle of radius 4

A = πR²/2 + πr²/2

A = π(12² + 4²)/2

A = 160π/2

A = 80π

8 0

2 years ago

The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh

mr_godi [17]

Answer:

a) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

b) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(10,0.15)$

Where $\mu=10$ and $\sigma=0.15$

We can calculate the probability of being defective like this:

$P(X$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

And if we replace we got:

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects.

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

Part c What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0

3 years ago

What is the relationship between a conjecture, a theorem, two column proof

Afina-wow [57]

Conjecture
is an educated guess based on known information. Reached by using inductive reasoning

4 0

3 years ago

-1(x+5)=3[x+(2x-1)]<br> solved and justified each step

madreJ [45]

Answer:

Step-by-step explanation:

$-1(x+5)=3[x+2x-1)]\\\\\text{for}\ -1(x+5):\ \text{distribtutive property}\\\text{for}\ [x+(2x-1)]:\ \text{associative property}\\\\(-1)(x)+(-1)(5)=3[(x+2x)-1]\\-x-5=3(3x-1)\\\\\text{for}\ 3(3x-1):\ \text{distributive property}\\\\-x-5=(3)(3x)+(3)(-1)\\-x-5=9x-3\\\\\text{for the equation}:\ \text{addition property of equality}\\\\-x-5=9x-3\qquad\text{add 5 to both sides}\\-x-5+5=9x-3+5\\-x=9x+2\\\\\text{for the equation:}\ \text{subtraction property of equality}\\\\-x=9x+2\qquad\text{subtract}\ 9x\ \text{from both sides}$

$-x-9x=9x-9x+2\\-10x=2\\\\\text{for the equation:}\ \text{division property of equality}\\\\-10x=2\qquad\text{divide both sides by -10}\\\dfrac{-10x}{-10}=\dfrac{2}{-10}\\x=-0.2$

$\text{Distributive property:}\ a(b+c)=ab+ac\\\text{Associative property:}\ (a+b)+c=a+(b+c)$

6 0

3 years ago