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Solnce55 [7]
3 years ago
7

Write a word problem whose solutions is |20| =20

Mathematics
1 answer:
Anton [14]3 years ago
5 0
You have 20 candies together in abag 10 red 10 green tou mix them yp togethr in a bag how many candies do you have in the bag
You might be interested in
We learned in Exercise 4.18 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample o
liraira [26]

Answer:

a) 108 people with a standard deviation of 3.286335

b) No

c) 0.218163 or around 21.82%

d) See explanation below.

Step-by-step explanation:

This situation can be modeled with the Binomial Distribution which gives t<em>he probability of an event that occurs exactly k times out of n, and is given by </em>

<em>\large P(k;n)=\binom{n}{k}p^kq^{n-k} </em>

<em>where  </em>

<em>\large \binom{n}{k}= combination of n elements taken k at a time. </em>

<em>p = probability that the event (“success”) occurs once </em>

<em>q = 1-p </em>

In this case, the event “success” is finding an American adult who had  chickenpox before adulthood with probability p=0.9

and n=120 American adults in the sample.

The <em>standard deviation for this binomial distribution</em> is

\large s=\sqrt{npq}

where <em>n is the sample size 120 </em>

a)

We consider a random sample of 120 American adults. How many people in this sample would you expect to have had chickenpox in their childhood?

If about 90% of American adults had chickenpox before adulthood, we expect to find 90% of 120 = 0.9*120=108 people in the sample who had chickenpox in their childhood.

The  standard deviation would be

\large s=\sqrt{120*0.9*0.1=3.286335}

b)

Would you be surprised if there were 105 people who have had chickenpox in their childhood?

No, because 105 and 108 are in the interval [mean - s, mean +s]

c)

What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?

The probability that 105 or fewer people in this sample have had chickenpox in their childhood is

\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}

We compute this value easily with a spreadsheet and we get

\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}=0.218163\approx 21.82\%

d)

How does this probability relate to your answer to part (b)?

A binomial distribution with np>5 and nq>5 with n the sample size as is the case here, behaves pretty much like a Normal distribution with mean np and standard deviation \large \sqrt{npq}, so around 60% of the data are in the interval  [mean -s, mean +s] and 40% outside, so roughly <em>20% of the data should be in [0, mean-s] </em>

4 0
3 years ago
For the vectors u and v with magnitudes u = 88 and v = 99​, find the angle θ between u and v which makes u and v = 77.
WARRIOR [948]

Answer:

<h2>89.5°</h2>

Step-by-step explanation:

Using the vector formula u.v = |u||v|cos\theta

|u| = magnitude of vector u

|v| = magnitude of vector v

u.v is the dot product of vector u and v

Given |u| = 88, |v| = 99 and u.v = 77, to get \theta we will substitute the given values into the equation above;

77 = 88*99cos\theta\\cos\theta = \frac{77}{88*99} \\cos\theta = \frac{77}{8712} \\cos\theta = 0.008838\\\theta = cos^{-1} 0.008838\\\theta = 89.5^{0}

4 0
3 years ago
The perimeter of a rectangle is 70 cm. The ratio of length to width is 2:5. Find the length and width of the rectangle.
olya-2409 [2.1K]

Answer:

length = 10 cm; width = 25 cm

Step-by-step explanation:

Let's call the length 2x and the width 5x. Since perimeter can be calculated by multiplying the sum of the length and width by 2 we can write:

2 * (2x + 5x) = 70

2 * (7x) = 70

7x = 35

x = 5 which means the length is 2 * 5 = 10 and the width is 5 * 5 = 25.

5 0
2 years ago
Can someone help me please I would appreciate it.
Dmitry_Shevchenko [17]
The answer is c to my calculations.
4 0
3 years ago
Udent
vitfil [10]

Answer:

B ; C ; D

Step-by-step explanation:

Number of faces on a number cube = 6

Sample space = (1, 2, 3, 4, 5, 6)

P(1 then 0)

P(1) = 1/6 ; P(0) = 0

P(1 then 0) = 1/6 * 0 = 0

P(even number then odd number) :

P(even number) = 3/6 = 1/2

P(odd) = 3/6 = 1/2

P(even number then odd number) = 1/2 * 1/2 = 1/4

P(6 then 2) :

P(6) = 1/6 ; P(2) = 1/6 = 1/2

P(6 then 2) = 1/6 * 1/6 = 1/36

P(even number then 5) :

P(even) = 3/6 = 1/2

P(5) = 1/6

P(even number then 5) = 1/2 * 1/6 = 1/12

P(odd number then 2) :

P(odd) = 3/6 = 1/2

P(2) = 1/6

P(odd number then 2) = 1/2 * 1/6 = 1/12

7 0
3 years ago
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