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3 years ago

Help me please.......

Mathematics

1 answer:

Tomtit [17]3 years ago

6 0

Answer:

29.4 %

Step-by-step explanation:

The area of the sector is directly proportional to the angle subtending the arc.

Area of sector/area of circle = 106°/360° = 0.294

The area of the sector is 29.4 % the area of the circle, so

A point chosen at random will be in the shaded region 29.4 % of the time.

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Drag equations to each row to show why the equation of a straight line is y = 3x, where 3 is the slope.

a_sh-v [17]

y/x = 3/1 Triangle 1 is similar to triangle 3

y = 3 x Multiply each side by x

4 0

3 years ago

Read 2 more answers

What is the measure of each interior angle in a regular 18-gon?

s344n2d4d5 [400]

<span>The answer is B) 160
First find exterior angle
360 / 18 = 20
Interior angle = 180 - 20 = 160

</span>

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3 years ago

Read 2 more answers

Arlene is comparing the prices of two truck rental companies. Company A charges $7 per hour and an additional $30 as service cha

Valentin [98]

Answer:

Part A: 7n + 30= -5 and 8n + 40= 16

Part B: 7(8) + 30= 86 and 8(8) + 40= 104

Part C: Company A saves you 12.

Step-by-step explanation:

Company A: 7n + 30

Company B: 8n + 40

7n + 30= -5 8n + 40= 16

7(8) +30= 86 8(8) +40= 104

7(2) + 30= 44 8(2) + 40= 56

56 - 44= 12

7 0

4 years ago

Is X-1 a factor of x^5-1?

Musya8 [376]

Answer: No

Explanation:

According to factor theorem, if f(x)=0 then x is a factor of the given function or equation.

As x-1 is a factor

We equate x-1=0

x=1

Substituting in x^5-1, we have 1^5-1 =1-1=0.

Hence, it's a factor.

When coming to x^5+1, it would become 1^5+1=1+1=2

So x-1 isn't a factor of x^5+1.

4 0

2 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago