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Debora [2.8K]
2 years ago
12

Initial Knowledge Check

Mathematics
1 answer:
Anna [14]2 years ago
5 0

Answer:

0

5

3

becos i know UwU

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The circumference of the hub cap of a tire is 83.21 centimeters. Find the area of this hub cap. Use
Elena L [17]

Answer:

<u>551 square centimeters</u>

Step-by-step explanation:

Finding the radius (r) :

  • C = 2πr
  • 83.21 = 2 x 3.14 x r
  • 83.21 = 6.28r
  • r = 83.21/6.28
  • r = 13.25 cm

Finding the area :

  • A = πr²
  • A = 3.14 x (13.25)²
  • A = 3.14 x 175.5625
  • A = <u>551 square centimeters</u> (nearest whole number)

<u></u>

The area of this hub cap is about <u>551 square centimeters</u>.

7 0
1 year ago
Read 2 more answers
Find the area of the square ABCD.
zmey [24]

Answer:

Step-by-step explanation:

area of square is basically one side times itself

14 times 14 =196m^2

4 0
2 years ago
Read 2 more answers
F(x) = -3(x - 1)2 + 3<br> How do you turn it into standard form
Zepler [3.9K]

Answer:

\displaystyle \boxed{f(x) = 6x - 3x^2}

Step-by-step explanation:

f(x) = −3(x² - 2x + 1) + 3

f(x) = −3x² + 6x - 3 + 3

f(x) = −3x² + 6x

** The above answer is written in reverse, which is the exact same result.

I am joyous to assist you anytime.

7 0
3 years ago
Help picture attached
Vilka [71]
I would say c or d. but since its multiple choice, its ok to guess sometimes.
6 0
3 years ago
A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective
Inessa05 [86]

Answer:

(a) P(X \leq 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....

where, n = number of samples taken = 150

            r = number of success

           p = probability of success which in our question is % of bulbs that

                  are defective, i.e. 10%

<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>

So, Let X = No. of defective bulbs in a box

<u>Mean of X</u>, \mu = n \times p = 150 \times 0.10 = 15

<u>Standard deviation of X</u>, \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.10 \times (1-0.10)} = 3.7

So, X ~ N(\mu = 15, \sigma^{2} = 3.7^{2})

Now, the z score probability distribution is given by;

                Z = \frac{X-\mu}{\sigma} ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X \leq 20) = P(X < 20.5)  ---- using continuity correction

    P(X < 20.5) = P( \frac{X-\mu}{\sigma} < \frac{20.5-15}{3.7} ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = n \times p = 150 \times 0.10 = 15.

                                           

5 0
3 years ago
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