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Mice21 [21]
3 years ago
9

assume that when adults with smartphones are randomly selected 15 use them in meetings or classes if 15 adult smartphones are ra

ndomly selected, find the probability that at least 4 of them use their smartphones
Mathematics
1 answer:
Tanzania [10]3 years ago
3 0

Answer:

The probability that at least 4 of them use their smartphones is 0.1773.

Step-by-step explanation:

We are given that when adults with smartphones are randomly selected 15% use them in meetings or classes.

Also, 15 adult smartphones are randomly selected.

Let X = <em>Number of adults who use their smartphones</em>

The above situation can be represented through the binomial distribution;

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

where, n = number of trials (samples) taken = 15 adult smartphones

           r = number of success = at least 4

           p = probability of success which in our question is the % of adults

                 who use them in meetings or classes, i.e. 15%.

So, X ~ Binom(n = 15, p = 0.15)

Now, the probability that at least 4 of them use their smartphones is given by = P(X \geq 4)

P(X \geq 4) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= 1- \binom{15}{0}\times 0.15^{0} \times (1-0.15)^{15-0}-\binom{15}{1}\times 0.15^{1} \times (1-0.15)^{15-1}-\binom{15}{2}\times 0.15^{2} \times (1-0.15)^{15-2}-\binom{15}{3}\times 0.15^{3} \times (1-0.15)^{15-3}

= 1- (1\times 1\times 0.85^{15})-(15\times 0.15^{1} \times 0.85^{14})-(105 \times 0.15^{2} \times 0.85^{13})-(455 \times 0.15^{3} \times 0.85^{12})

= <u>0.1773</u>

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The difference between two positive integers is 7 and the sum of their squares is 949. What are the numbers?
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Answer:

25 and 18

Step-by-step explanation:

Let's say that the first number is x and the second one is y.

First, the difference between them is 7, so x-y=7

Next, the sum of their squares is 949, so x²+y² = 949

We have

x-y=7

x²+y²=949

One thing we can do to solve this problem is to solve for x in the first equation, plug that into the second equation, and go from there

Adding y to both sides in the first equation, we have

x = 7 + y

Plugging that into the second equation for x, we have

(7+y)²+ y² = 949

expand

(7+y)(7+y) + y² = 949

49 + y² + 7y + 7y + y² = 949

combine like terms

2y² +14y + 49 = 949

subtract 949 from both sides to put this in the form of a quadratic equation

2y² + 14y - 900 = 0

divide both sides by 2

y² + 7y - 450 = 0

To factor this, we want to find 2 numbers that add up to 7 and multiply to -450.

The factors of 450 are as follows:

1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 30, 45, 50, 75, 90, 150, 225, and 450.

Note that we want to find two numbers with a difference of 7, as one will have to be negative for the multiplication to end up at -450. Two numbers that stand out are 18 and 25. To make them add up to 7, 18 can be negative. We therefore have

y² + 25y - 18y - 450 = 0

y(y+25) - 18(y+25) = 0

(y-18)(y+25) = 0

Solving for 0,

y-18 = 0

add 18 to both sides

y=18

y+25 = 0

subtract 25 from both sides

y= -25

As the question states "two positive integers", this means that y must be positive, so y = 18. We know x-y=7, so

x-18 = 7

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The marketing club at school is opening a student store. They randomly survey 50 students about how much money they spend on lun
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The expected value for a student to spend on lunch each day = $5.18

Step-by-step explanation:

Given the data:

Number of students______$ spent

2 students______________$10

1 student________________$8

12 students______________$6

23 students______________$5

8 students_______________$4

4 students_______________$3

Sample size, n = 50.

Let's first find the value on each amount spent with the formula:

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For $8:

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To find the expected value a student spends on lunch each day, let's add all the values together.

Expected value =

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Answer:

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8x = - 4(x + 3)

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+4x  +4x

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12       12

x = -1

  • Remember, a negative times a positive equals a negative
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So, the capacity of x buses  = 40(x)  = 40 x

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So, the capacity of y buses  = 50(y)  = 50 y

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Also, the total number of drivers available  = 9

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Also, x  ≤ 8,   y ≤ 10

Now, solving both equations, we get:

40 x + 50 y = 400 ..... (1)

x +  y = 9  ⇒ y = (9-x) put in (1)

40 x + 50 y = 400  ⇒  40 x  + 50 (9-x)  = 400

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or, - 10 x  =- 50

or, x  = 5 ⇒ y = (9-x)  = 9- 5 = 4

Hence the number of small buses used = 5

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