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Alekssandra [29.7K]
2 years ago
7

HELP QUICK! NO BOTS!

Mathematics
1 answer:
gtnhenbr [62]2 years ago
6 0

Answer:I'm pretty sure it's A

Step-by-step explanation:

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Which value is equivalent to cos 10°?<br> sin 30°<br> O sin 25°<br> O sin 70°<br> O sin 80°
Sauron [17]

Answer:

13

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
PLEASE HELP ME WITH BOTH QUESTIONS<br> THANKS!!
vitfil [10]

First one:

cos(A)=AC/AB=3/4.24

cos(B)=BC/AB=3/4.24

Cos(A)/cos(B)=AC/AB / (BC/AB) = AC/AB * AB/BC = AC/BC=3/3=1


Second one:

To solve this problem, we have to ASSUME AFE is a straight line, i.e. angle EFB is 90 degrees. (this is not explicitly given).

If that's the case, AE is a transversal of parallel lines AB and DE.

And Angle A is congruent to angle E (alternate interior angles).

Therefore sin(A)=sin(E)=0.5

8 0
3 years ago
Simplify
OlgaM077 [116]

Answer:

see below

Step-by-step explanation:

\frac{x-1}{x^2-3x+2}+ \frac{x-2}{x^2-5x+6} +\frac{x-5}{x^2-8x+15}

we need to simplify that

x^2-3x+2=(x-1)(x-2)\\\\x^2-5x+6=(x-2)(x-3)\\\\x^2-8x+15=(x-3)(x-5)

so we can continue

\frac{x-1}{(x-1)(x-2)}=\frac{1}{x-2}\\\\\frac{x-2}{(x-2)(x-3)} =\frac{1}{x-3}\\\\\frac{x-5}{(x-3)(x-5)} =\frac{1}{x-3}

and we can put all together

\frac{1}{x-2}+ \frac{1}{x-3}+ \frac{1}{x-3}\\\\\frac{1}{x-2} +\frac{2}{x-3}\\\\\frac{x-3}{(x-3)(x-2)}+ \frac{2(x-2)}{(x-2)(x-3)} \\\\\frac{x-3+2x-4}{(x-3)(x-2)}\\\\\frac{3x-7}{x^2-5x+6}

3 0
3 years ago
The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a m
Mazyrski [523]

Answer:

The probability that the instrument does not fail in an 8-hour shift is P(X=0) \approx 0.8659

The probability of at least 1 failure in a 24-hour day is P(X\geq 1 )\approx 0.3508

Step-by-step explanation:

The probability distribution of a Poisson random variable X representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:

P(X)=\frac{e^{-\mu}\mu^x}{x!}

Let X be the number of failures of a testing instrument.

We know that the mean \mu = 0.018 failures per hour.

(a) To find the probability that the instrument does not fail in an 8-hour shift, you need to:

For an 8-hour shift, the mean is \mu=8\cdot 0.018=0.144

P(X=0)=\frac{e^{-0.144}0.144^0}{0!}\\\\P(X=0) \approx 0.8659

(b) To find the probability of at least 1 failure in a 24-hour day, you need to:

For a 24-hour day, the mean is \mu=24\cdot 0.018=0.432

P(X\geq 1 )=1-P(X=0)\\\\P(X\geq 1 )=1-\frac{e^{-0.432}0.432^0}{0!}\\\\P(X\geq 1 )\approx 0.3508

3 0
3 years ago
I need help with question 3, I chose the answers that made the most sense but it keeps telling me I'm wrong?
Zarrin [17]

Answer: DL, PS, SL

Step-by-step explanation:

All sides of a square are congruent.

8 0
2 years ago
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