Ask question

Ask question

All categories

2 years ago

9

Kamden is a mind-reading alien. He had each friend draw and look at a card, then he guessed whether the card had a star on it. T

he two-way frequency table below shows data on guess and true status of the cards the friends drew.

1 answer:

torisob [31]2 years ago

7 0

Answer: This is what Khan Academy said the answer was

You might be interested in

If 1 japanese yen equals 0.0079 euros, then 1 euro equals approximately how many japanese yen?

Oliga [24]

We know that

<span>applying a rule of three</span>
if <span>1 japanese yen -------------> equals 0.0079 euros
X </span> japanese yen--------------> 1 euro
x=[1 euro]*[1 japanese yen]/[0.0079 euros]------------> 126.58 japanese yen

the answer is
126.58 japanese yen

6 0

2 years ago

The base of a solid right pyramid is a regular hexagon with

cupoosta [38]

Answer: Last option.

Step-by-step explanation:

Observe the base of the pyramid, which is a regular hexagon. You can see that there is a right triangle.

The area of the triangle can be calculated with this formula:

$A_t=\frac{bh}{2}$

Where "b" is the base and "h" is the height.

You can say that the apothem is the height of the right triangle, then, you need to find the base applying the Pythagorean Theorem:

$(2x)^2=(x\sqrt{3})^2+b^2\\\\b=\sqrt{(2x)^2-(x\sqrt{3})^2} \\\\b=\sqrt{4x^2-3x^2}\\\\b=\sqrt{x^2}\\\\b=x$

Then, the area of the triangle is:

$A_t=\frac{x(x\sqrt{3})}{2}=\frac{x^2\sqrt{3})}{2}$

Since the base of the pyramid is a regular hexagon, then you can multiply by 12 the area of the right triangle calculated above, in order to find the area of the hexagon. Then you get:

$A_h=12(\frac{x^2\sqrt{3})}{2})=6x^2\sqrt{3}\ units^2$

5 0

2 years ago

Read 2 more answers

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

5 0

2 years ago

Avery wants to put a variety of muffins in a display case. The case is large enough to hold 60 muffins. types of muffins Blueber

IgorLugansk [536]

There are 54 muffins in the basket al you have to do is multiply the 60 by the added percentages which is .9 and you get the answer of 54

3 0

2 years ago

Last year your professor wanted to study the pattern in class grades. The following table contains a sample of grades collected

Maslowich

Answer:

mean = 78.4

median = 77.5

mode = 75

This is Right - skewed (positive skewness) distribution

Step-by-step explanation:

<u>Mean:-</u>

The mean (average) is found by adding all of the numbers together and dividing by the number of items and it is denoted by x⁻

mean = $\frac{64+80+75+98+75}{5}$

mean (x⁻ ) = 78.4

The mean of the given data = 78.4

<u>Median:</u>

The median is found by ordering the set from lowest to highest and finding the exact middle.

64 ,75, 80, 98

The middle term of the given data set = $\frac{75+80}{2} =77.5$

<u>Mode :</u>

The mode is the most common repeated number in a data set.

64 ,75, 75, 80, 98

in data the most common number = 75

<u>Conclusion</u>:-

mean = 78.4

median = 77.5

mode = 75

This is Right - skewed (positive skewness) distribution

8 0

2 years ago