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kotegsom [21]
3 years ago
11

Help me please I don’t have much time

Mathematics
1 answer:
eduard3 years ago
4 0

Answer:

A.

Step-by-step explanation:

Where d is the difference

a_{n} = a_{1} + (n - 1) d

a_{n} = a_{1} + (n - 1) 7

(Because 8 - 1 is 7)

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A public health organization reports that 40%of baby boys 6-8 months old in the United
Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

  • 0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.
  • 0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.
  • Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

P(X > 4) = 1 - P(X \leq 4)

In which:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061

P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403

P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150

P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502

Hence:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325

P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106

P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016

P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

4 0
1 year ago
A drawer contains 8 red shirts, 6 blue shirts, and 7 white shirts. If a shirt is drawn at random, what is the probability that t
zzz [600]
7 over 21 . because you add 7 plus 8 plus 6 and get 21 then have 7 white shirts 
8 0
3 years ago
Read 2 more answers
(pls help me with this)
exis [7]

A. The coordinates of the midpoint of CD in terms of p and q is [(4 + p) / 2 , (5 + q) / 2]

B. The coordinates of D, Given that the midpoint of CD is (7, 1) is (10 , -3)

<h3>A. How to determine the mid point</h3>
  • Coordinate of C = (4, 5)
  • Coordinate of D = (p, q)
  • Mid point =?

Mid point = (X , Y)

X = (x₁ + x₂) / 2

X = (4 + p) / 2

Y = (y₁ + y₂) / 2

Y = (5 + q) / 2

Thus,

Mid point = (X , Y)

Mid point = [(4 + p) / 2 , (5 + q) / 2]

<h3>B. How to determine the coordinates of D</h3>
  • Mid point = (7, 1)
  • Coordinates of D =?

Mid point = (7, 1) = (X , Y)

X = (4 + p) / 2

7 = (4 + p) / 2

Cross multiply

7 × 2 = 4 + p

14 = 4 + p

Collect like terms

p = 14 - 4

p = 10

Y = (5 + q) / 2

1 = (5 + q) / 2

Cross multiply

1 × 2 = 5 + q

2 = 5 + q

Collect like terms

q = 2 - 5

q = -3

Coordinates of D = (p, q)

Coordinates of D = (10 , -3)

Learn more about coordinate geometry:

brainly.com/question/4976351

#SPJ1

5 0
2 years ago
1. A weighted coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 4 times, what is the pr
ohaa [14]

Answer:

80/81

Step-by-step explanation:

If a head is twice as likely to occur as a tail, then the probability of getting heads is 2/3 and the probability of getting tails is 1/3.

The probability of getting at least 1 head involves 4 scenarios:

1) 1 Head and 3 Tails

2) 2 Heads and 2 Tails

3) 3 Heads and 1 Tail

4) 4 Heads

Instead of calculate all these scenarios, you could calculate the opposite scenario: 4 Tails. The sum of all possible scenarios is 1, so:

P(at least one head) + P(no heads) = 1

Then, P(at least one head) = 1 - P(no heads)

The probability of 4 tails is:

P(no heads) = P(TTTT) = (1/3)(1/3)(1/3)(1/3)=1/81

Then, P(at least one head) = 1 - 1/81=80/81

7 0
3 years ago
CAN SOMEONE PLEASE HELP ME???<br> ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
OLEGan [10]

Answer:

Option D, AP

Step-by-step explanation:

<u>The two sides that are connected by the right angle are called the sides.  The longest one that is not connect with a right angle is called the hypotenuse.</u>

<u />

In this case, the sides are AZ and PZ.  <em>The hypotenuse is AP</em>

<em />

Answer:  Option D, AP

6 0
3 years ago
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