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kondaur [170]
3 years ago
6

Select the graph of the equation below. y= -1/4x^2+1

Mathematics
1 answer:
V125BC [204]3 years ago
4 0

Answer:

a.

Step-by-step explanation:

Since you have the different parabolas there, the simplest way to determine which one is, is replacing x as 0 and then y as 0

y = -1/4x^2 + 1

When x = 0

y = -1/4 (0)^2 +1 = 1

When y = 0

0 = -1/4x^2 +1

1/4 x^2 = 1

x^2 = 4

x = + 2

x = - 2

The parabola with points:

(0,1); (-2,0) and (2,0) is a.

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The answer is 45°

By determining the angles of triangle ACB (A=30°, B=45°, C=105° (total of 180°)) and knowing that angle 1 correlates to angle 4 ((1)°=(4)°), we can find angle 1, which is 45°, equal to angle 4.

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Given two circles whose radii are 5" and 15", the ratio of their areas is
Serhud [2]

Answer:

The answer to your question is ratio = 9

Step-by-step explanation:

Data

Circle 1 radius = 5''

Circle 2 radius = 15''

Process

1.- Calculate the area of circle 1

Area = πr²

Area = π(5)²

Area = 25π

2.- Calculate the area of circle 2

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Area = π(15)²

Area = 225π

3.- Calculate the ratio of the areas

ratio = 225π / 25π  

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6 0
3 years ago
Find the probability that in five tosses of a fair die, a 3 will appear (a) twice, (b) at most once, (c) at least two times.
Jobisdone [24]

Answer:

A) 0.1612

B) 0.8031

C) 0.1969

Step-by-step explanation:

For each toss of the die, there are only two possible outcomes. Either it is a 3, or it is not. The probability of getting a 3 on each toss is independent from other tosses. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

Five tosses, so n = 5

The die has 6 values, from 1 to 6. The die is fair, so each outcome is equally as likely. The probability of a 3 appearing in a single throw is p = \frac{1}{6} = 0.167

(a) twice

This is P(X = 2)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{5,2}.(0.167)^{2}.(0.833)^{3} = 0.1612

(b) at most once

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

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.167)^{0}.(0.833)^{5} = 0.4011

P(X = 1) = C_{5,1}.(0.167)^{1}.(0.833)^{1} = 0.4020

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.4011 + 0.4020 = 0.8031

(c) at least two times.

Either a 3 appears at most once, or it does at least two times. The sum of the probabilities of these events is decimal 1. So

P(X \leq 1) + P(X \geq 2) = 1

We want P(X \geq 2)

So

P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.8031 = 0.1969

6 0
3 years ago
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