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dem82 [27]
3 years ago
13

One number is one less than five times another. if their sum is decreased by three, the result is eight. find the numbers.

Mathematics
1 answer:
Anestetic [448]3 years ago
5 0

<span>One number is one less than five times another ....
</span>Let one number be x<span>
The other number is 5x - 1

Their sum ... 
x + 5x - 1 = 6x - 1

Their sum is decreased by 3
6x - 1 - 3 = 6x - 4

The result is eight
6x - 4 = 8
6x = 8 + 4
6x = 12
x = 2

x = 2
</span>5x - 1 = 5(2) - 1 = 9<span>
One number is 2 and the other is 9.</span>
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The answer is 3/5 because u need to find the missing side in order to find sin

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3 years ago
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What is the length of the tangent line in the picture?
S_A_V [24]

Answer:

x = -2

Step-by-step explanation:

The tangent line has length "x + 8"

The secant line has length "x + 6 + 5", where

5 is the inner part

x + 6 is the outer part

Now,

the secant-tangent theorem tells us that square of the tangent line is equal to the outer segment of secant line multiplied by length of whole secant line.

So, we can say:

(x+8)^2 = (x+6)(x+6+5)

We can solve for x shown below:

(x+8)^2 = (x+6)(x+6+5)\\(x+8)^2=(x+6)(x+11)\\x^2+16x+64=x^2+17x+66\\17x-16x=64-66\\x=-2

The value of x is -2

4 0
3 years ago
The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(
forsale [732]

Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

Considering that P(BIA)= \frac{P(AnB)}{P(A)} you can clear the intersection from the formula P(AnB)= P(B/A)*P(A) and apply it for the given events:

P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07

P(A_2nB)= P(B/A_2)*P(A_2)= 0.05*0.50= 0.025

c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74

P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26

I hope it helps!

5 0
3 years ago
Find the values of x and a please!!
Ludmilka [50]

Answer:

yes

Step-by-step explanation:

6 0
3 years ago
100 POINTS
likoan [24]

Answer: True because when you divide both sides by 2 that would be the right equation.

Step-by-step explanation:

3 0
3 years ago
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