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2 years ago

Find the circumference of a circle with a diameter of 12 inches. Round to the nearest

Mathematics

1 answer:

Andru [333]2 years ago

4 0

Answer:

~37.70 inches

Step-by-step explanation:

The circumference is π x diameter, so if the diameter is 12, the circumference is π x 12=12π

12π is approximately 37.70

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The measure of an angle is 12°. What is the measure of its supplement?<br> *No answer just number*

Elena-2011 [213]

Answer:

168°

Step-by-step explanation:

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2 years ago

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What is the slope of the line that passes through the points (1,0) and (5, 3)?

Varvara68 [4.7K]

Answer:

Undefined

Step-by-step explanation:

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2 years ago

Consider randomly selecting a student at a large university. Let A be the event that the selected student has a Visa card, let B

wariber [46]

Answer:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) is the probability that given that a student has a visa card, they also have a master card

p(A/B) is the probability that given a student has a master card, they also have a visa card

e. 0.35

f. 0.31

Step-by-step explanation:

a. p(AUBUC)= P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC)

=0.6+0.4+0.2-0.3-0.11-0.1+0.07= 0.76

b. P(AnBnC')= P(AnB)-P(AnBnC)

=0.3-0.07= 0.23

c. P(B/A)= P(AnB)/P(A)

=0.3/O.6= 0.5

e. P((AnB)/C))= P((AnB)nC)/P(C)

=P(AnBnC)/P(C)

=0.07/0.2= 0.35

f. P((AUB)/C)= P((AUB)nC)/P(C)

=(P(AnC) U P(BnC))/P(C)

=(0.11+0.1)/0.2

=0.21/0.2 = 0.31

7 0

2 years ago

The scale on a map says that 6 cm represents 15 km. What distance on the map in centimeters represents 20 kilometers?

ale4655 [162]

Answer:

8 centimeters.

Step-by-step explanation:

$6cm\ on\ map\ -> 15km\\$

Divide 3 both sides:

$2cm\ on\ map\ -> 5km$

Multiply 4 on both sides:

$8cm\ on\ map\ -> 20km$

8cm on the map represents 20km.

4 0

2 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

2 years ago