Each cupcake costs $4.00 how many cupcakes (x) are purchased if the total cost is $36.00

Rework problem 28 from section 3.1 of your text, involving the inspection of refrigerators on an assembly line. You should still

sergeinik [125]

Event: Probability: A. Too much enamel 0.18 B. Too little enamel 0.24 C. Uneven application 0.33 D. No defects noted 0.47

let P(AC) = x, P(BC) = y, then P(A) + P(B) + P(C) - (x+y) = 1-0.47 = 0.53 x+y = 0.22
3. The probability of paint defects that results to an improper amount of paint and uneven application? 
P(A U B U C) = 0.53

4. the probability of a paint defect that results to
the proper amount of paint, but uneven application?
P(C) - P(AC) - P(BC) = 0.47 - 0.22 = 0.25

A and B are disjoint so P(ABC) = 0, but you can have P(AC) and P(BC). you can't compute these separately here, but you can compute P(AC) + P(BC). By the way, P(AC) eg is just an abbreviated version of P(A∩C).

3 0

2 years ago

Susan plans to use 120 feet of fencing to enclose a rectangular area for a garden. Which equation best models the area, y, of th

mina [271]

Well we set the perimeter to 120 feet.
This means that 2x+2y=120
Now we know the area of a rectangle is xy so we have to solve for both x and y in the perimeter equation.

2x=120-2y
x=60-y

2y=120-2x
y=60-x

Now we plug these values into our area equation A=xy to get:
A=(60-y)(60-x)

7 0

2 years ago

Read 2 more answers

One of the vertices of △PQR is P(2, −1). The midpoint of PQ is M(3, 0). The midpoint of QR is N(5, 3). Show that MN || PR and MN

VLD [36.1K]

Answer:

See the proof below

Step-by-step explanation:

Given the following coordinates

P(2, −1)

Midpoint of PQ M(3, 0)

We can get the coordinate point Q using the midpoint formula;

M(X,Y) = (x1+x2/2, y1+y2/2)

X = x1+x2/2

3 = 2+x2/2

6 = 2+x2

x2 = 6-2

x2 = 4

Y = y1+y2/2

0 = -1+y2/2

0 = -1 + y2

y2 = 0+1

y2 = 1

Hence the coordinate of Q is (4, 1)

Next is to get the coordinate of R

Given the midpoint of QR to be N(5, 3)

(5,3) = (4+x2/2, 1+y2/2)

5 = 4+x2/2

10 = 4+x2

x2 = 10-4

x2 = 6

1+y2/2 = 3

1+y2 = 6

y2 = 6-1

y2 = 5

Hence the coordinate of R is (6,5)

Given the coordinates M(3, 0) and N(5, 3)

Slope is expressed as:

m = y2-y1/x2-x1

m = 3-0/5-3

m = 3/2

Slope of MN = 3/2

Get the slope of PR

Given the coordinates P(2, −1) and R (6,5)

Slope of PR = 5-(-1)/6-2

Slope of PR = 5+1/4

Slope of PR = 6/4 = 3/2

Since the slope of MN is equal to that of PR, hence MN is parallel to PR i.e MN || PR

To show that MN = 1/2PR, we will have to take the distance between M and N and also P and R first as shown:

For MN with coordinates M(3, 0) and N(5, 3)

MN = √(x2-x1)²+(y2-y1)²

MN = √(5-3)²+(3-0)²

MN = √2²+3²

MN = √13

Get the length of PR where P(2, −1) and R (6,5)

PR = √(6-2)²+(5+1)²

PR = √4²+6²

PR = √16+36

PR = √52

PR = √4*13

PR = √4*√13

PR = 2√13

Since MN = √13

PR = 2MN

Divide both sides by 2

PR/2 = 2MN/2

PR/2 = MN

Hence MN = 1/2 PR (Proved!)

8 0

1 year ago

An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t

Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .