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gtnhenbr [62]
3 years ago
15

Please help me with this problem

Mathematics
1 answer:
victus00 [196]3 years ago
5 0

Answer:

r= 8

Step-by-step explanation:

Devide both sides by -4

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Solve the following equation. Determine if the equation is an "identity" or a "contradiction."
Eva8 [605]
-4t + 1 = 3t + 1 -7t\\
-4t+1=-4t+1

It's identity.
4 0
4 years ago
What is the general equation of a sine function with an amplitude of 6, a period of StartFraction pi Over 4 EndFraction, and a h
Katen [24]

Answer:

C: y = 6 \sin (8(x-\frac{\pi}{2} ))

Step-by-step explanation:

Amplitude is changed by multiplying the entire trig function by a number, by default the amplitude is 1, so multiplying by 6 is necessary; eliminate A and B

Horizontal shift is going through the x axis, so it must be subtracted from the x value. This is only seen in choice C of the remaining answers.

Calculating the period isn't necessary because we already found the right answer but in case you need it in the future; the period is by default 2pi, to get to pi/4, it had to be 8 times smaller, so multiplying x and translations on x by 8 will get you the correct period.

3 0
3 years ago
Read 2 more answers
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
Heyyy anyone wanna be friends
Mrac [35]

Answer:

ytes ofccc

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is 5(a+1)=2a+14
GenaCL600 [577]
If you are solving for a: it equals 3
distribute: 5a + 5=2a +14
get rid of extra number on one side: 5a=2a +9      (subtract 5 from both sides)
get rid of the other variable: 3a=9      (subtract 2a from both sides)
divide to isolate a: 3a/3=9/3      (divide by three on both sides)
answer= a=3

5 0
3 years ago
Read 2 more answers
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