A common denominator for 6 and 8 is

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%

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.

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

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 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

2 years ago

What is the line's slope

ra1l [238]

Answer:

It tell the direction and the steepness of the line. Which means how much they rise or how much they fall, horizontal or vertical. It represent by the letter m

Step-by-step explanation:

m= y/x

more detail

m= y₂ - y₁ / x₂ - x₁

4 0

3 years ago

Exponential growth can anyone give the answers to me? I’m completely lost!

PilotLPTM [1.2K]

Answer:

Q1

The exponential growth model for frogs:

F(x) = 100(1.22)ˣ,

F- number of frogs, x- number of years, 1.22 - growth factor

Calculations:

F(5) = 100(1.22)⁵ = 270
a) F(10) = 100(1.22)¹⁰ = 730
b) F(20) = 100(1.22)²⁰ = 5335

All numbers rounded

Q2

The exponential growth model for bacteria:

B(x) = 10(1.8)ˣ

B- number of bacteria, x- number of hours, 1.8 - growth factor

Calculations:

a) B(5) = 10(1.8)⁵ = 189
b) B(24) = 10(1.8)²⁴ = 13382588
c) B(168) = 10(1.8)¹⁶⁸ = 7.68 * 10⁴³

All numbers rounded

Q3.

The exponential growth model for fish:

F(x) = 821(1.02)ˣ,

F- number of fish, x- number of months, 1.02 - growth factor

Calculations:

a) F(12) = 821(1.02)¹² = 1041
b) F(120) = 821(1.02)¹²⁰ = 8838

All numbers rounded

3 0

2 years ago

Read 2 more answers

The point A (5,11) is reflected across the x-axis. How long is the segment AA'?

Nikitich [7]

Answer: The required length of the segment AA' is 11 units.

Step-by-step explanation: Given that the point A(5, 11) is reflected across the X-axis.

We are to find the length of the segment AA'.

We know that

if a point (x, y) is reflected across X-axis, then its co-ordinates becomes (x, -y).

So, after reflection, the co-ordinates of the point A(5, 11) becomes A'(5, -11).

Now, we have the following distance formula :

The DISTANCE between two points P(a, b) and Q(c, d) gives the length of the segment PQ as follows :

$PQ=\sqrt{(c-a)^2+(d-b)^2}.$

Therefore, the length of the segment AA' is given by

$AA'=\sqrt{(5-5)^2+(-11-11)^2}=\sqrt{11^2}=11.$

Thus, the required length of the segment AA' is 11 units.

7 0

2 years ago

8.3 is equal to _________ tenths.

SIZIF [17.4K]

Answer:

830 tenths

Step-by-step explanation:

8.0 ,equal to 800 tenths

0.3,equal to 30 tenths

8 0

2 years ago