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3 years ago

Help needed ASAP will give brainliest.

Mathematics

1 answer:

ValentinkaMS [17]3 years ago

7 0

Answer: Hi!

First, we should find the square root of 68, which is 8.25. (an easy way to find the square root for numbers like this is to just use the calculator, as we can't automatically tell what it is; 68 isn't a perfect square number.) Since we know know the square root we can easily evaluate the questions!

Our answer is c), since the points only span from 0 - 8 and the square root of 68 is 8.25. Since 8.25 is greater than 8, it would be outside of the points and to the right of point Q.

Hope this helps!

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Will give brqinlist if correct

BARSIC [14]

Answer:

7 + 8x

Step-by-step explanation:

-11 + 2 (4x -1) + 6

=> -11 + 8x -2 + 6

=> -13 + 8x + 6

=> -7 + 8x

So, the simplified expression is 7 + 8x.

5 0

3 years ago

ASAP!!!!!!!!! PLEASE help me with this question! This is really urgent! No nonsense answers please.

Vesna [10]

Answer:

Because <CBD is an inscribed angle and <CAD is a central angle with the same intercepted arc, m<CBD = 55°, or half of the measure of <CAD.

Step-by-step explanation:

The Inscribed Angle Theorem proves that an inscribed angle is half the measure of a central angle, if both the inscribed angle and the central angle intercepts the same arc.

Also, according to the inscribed angle theorem, an inscribed angle is ½ of the measure of the arc it intercepts.

Therefore, m<CBD is half of m<CAD, or half of the measure of the arc CD that they both intercept together.

Thus, m<CBD = 55°, which is ½ of m<arc CD.

m<arc CD = 110° = m<CAD.

m<CBD = ½ of m<CAD = 55°.

The statement that best describes the relationship between <CBD and <CAD is "Because <CBD is an inscribed angle and <CAD is a central angle with the same intercepted arc, m<CBD = 55°, or half of the measure of <CAD."

4 0

3 years ago

The diameter of a circular table is 6 feet. Which of the following is closest to the area of the tabletop

Ivahew [28]

So if you do 6(pi) then you should get 18.84

6 0

3 years ago

Read 2 more answers

Consider the recursive function,<br> f(1) = 2<br> f(n) = 5•f(n − 1), for n > 2

pogonyaev

Answer:

yes?

Step-by-step explanation:

??? can u say exactly what the question is please? thank you

7 0

3 years ago

Read 2 more answers

In a widget factory, machines A, B, and C manufacture, respectively, 20, 30, and 50 percent of the total. Of their output 6, 3,

steposvetlana [31]

Answer:

38.71% probability it was manufactured by machine A.

29.03% probability it was manufactured by machine B.

32.26% probability it was manufactured by machine C.

Step-by-step explanation:

We have these following probabilities:

A 20% probability that the chip was fabricated by machine A.

A 30% probability that the chip was fabricated by machine B.

A 50% probability that the chip was fabricated by machine C.

A 6% probability that a chip fabricated by machine A was defective.

A 3% probability that a chip fabricated by machine B was defective.

A 2% probability that a chip fabricated by machine C was defective.

The question can be formulated as:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

What are the probabilities that it was manufactured by machines A, B, and C?

Machine A

What is the probability that the widget was manufactures by machine A, given that it is defective?

P(B) is the probability that it was manufactures by machine A. So $P(B) = 0.20$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine A. So $P(A/B) = 0.06$

P(A) is the probability that a widget is defective. This is the sum of 6% of 20%, 3% of 30% and 2% of 50%. So

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.2*0.06}{0.031} = 0.3871$

38.71% probability it was manufactured by machine A.

Machine B

What is the probability that the widget was manufactures by machine B, given that it is defective?

P(B) is the probability that it was manufactures by machine B. So $P(B) = 0.30$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine B. So $P(A/B) = 0.03$

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.03}{0.031} = 0.2903$

29.03% probability it was manufactured by machine B.

Machine C

What is the probability that the widget was manufactures by machine C, given that it is defective?

P(B) is the probability that it was manufactures by machine C. So $P(B) = 0.50$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine C. So $P(A/B) = 0.02$

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.02}{0.031} = 0.3226$

32.26% probability it was manufactured by machine C.

6 0

3 years ago