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3 years ago

Which relation is a function?

1 answer:

muminat3 years ago

4 0

C. (8,-2), (-2,5), (6,0) is a function, because all the numbers are different unlike A, B, and D those all have the same number either in there domain or range.

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A tree is leaning toward the ground. How many degrees does the tree have to fall before hitting the ground?

Firdavs [7]

Im not absolutely sure but i think its 14

6 0

4 years ago

Read 2 more answers

Until 2014, the mass of a kilogram was defined as exactly equal to the mass of a special

emmainna [20.7K]

Step-by-step explanation:

We have,

Diameter of cylinder, d = 39.2 mm

Radius, r = 19.6 cm

Height of the cylinder, h = 39.2 mm

It is required to find the surface area of this cylinder. The formula of the surface area of cylinder is given by :

$S=2\pi rh+2\pi r^2\\\\S=2\pi r(h+r)$

Putting all values we get :

$S=2\times\dfrac{22}{7}\times19.6\cdot(39.2+19.6)\\\\S=7244.16\ mm^2$

7 0

3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago

-Solve the proportion (round to the nearest tenth). 1910=x12

zysi [14]

The value of x in the proportion is 15.8.

1910 = x12

1910 = 12x

divide both sides by 12

1910 / 12 = 12x / 12

x = 15.8333333333

x ≈ 15.8

5 0

3 years ago

A man was traveling by air is allowed a maximum of 20kg luggages .The man has four bags weighing 3.5kg ,15 kg ,2kg and 1.5kg .Fi

KatRina [158]

Answer:

See explanation

Step-by-step explanation:

Maximum weight = 20 kg

Bag 1 = 3.5kg

Bag 2 = 15 kg

Bag 3 = 2kg

Bag 4 = 1.5kg

Total weight of bags = 22 kg

Excess weight of his luggage = Total weight of bags - Maximum weight

= 22 kg - 20 kg

= 2 kg

Express the excess weight as a percentage of his maximum weight allowed = excess weight / maximum weight × 100

= 2/20 × 100

= 0.1 × 100

= 10%

Express the excess weight as a percentage of his maximum weight allowed = 10%

5 0

3 years ago