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3 years ago

:( help me please,,, if ur good at mean, median mode and range

Mathematics

1 answer:

Verdich [7]3 years ago

4 0

Answer:

ill help when you give the problem lol

Step-by-step explanation:

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It takes 70 inches of ribbon to make a bow and wrap the ribbon around a box. The bow takes 32 inches of ribbon. The width of the

Westkost [7]

You need two sides so you can rewrite it as:
2x = 70-(14+32)
2x = 24
x = 24/2
x = 12

The height is 12in.

Hope this helps! :)

7 0

3 years ago

is picking out some movies to rent, and he has narrowed down his selections to 5 documentaries, 7 comedies, 4 mysteries, and 5 h

pogonyaev

Answer: 91

Step-by-step explanation:

Given : The number of documentaries = 5

The number of comedies = 7

The number of mysteries = 4

The number of horror films =5

The total number of movies other than comedy = 14

Now, the number of possible combinations of 9 movies can he rent if he wants all 7 comedies is given by :-

$^7C_7\times^{14}C_2\\\\\dfrac{7!}{7!(7-7)!}\times\dfrac{14!}{2!(14-2)!}\\\\=(1)\times\dfrac{14\times13}{2}\\\\=91$

Therefore, the number of possible combinations of 9 movies can he rent if he wants all 7 comedies is 91 .

5 0

4 years ago

Pranked haha hkjkdvhpihigiuaspihgpiaisuhgiuasi free redit

iren2701 [21]

Okay the answer is 69 because 81-2 is 69

6 0

3 years ago

Read 2 more answers

Please help with three through 12

lord [1]

The correct answer for 12 is B

6 0

3 years ago

Read 2 more answers

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

3 years ago