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3 years ago

In a packet of 40 skittles,30% are red .How Many are not red?

Mathematics

1 answer:

elena55 [62]3 years ago

6 0

Answer:

28 skittles

Step-by-step explanation:

30% of 40 is 12

subtract 12 from 40 and that is 28

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Each child receives a meal which consists a hamburger a bag of pretzels and a drink for the children 6 -10 years old and a hambu

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Who many children is there total?

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4 years ago

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a

kipiarov [429]

Answer:

a) P(k≤11) = 0.021

b) P(k>23) = 0.213

c) P(11≤k≤23) = 0.777

P(11<k<23) = 0.699

d) P(15<k<25)=0.687

Step-by-step explanation:

a) What is the probability that the number of drivers will be at most 11?

We have to calculate P(k≤11)

$P(k\leq11)=\sum_0^{11} P(k$

$P(k=0) = 20^0e^{-20}/0!=1 \cdot 0.00000000206/1=0\\\\P(k=1) = 20^1e^{-20}/1!=20 \cdot 0.00000000206/1=0\\\\P(k=2) = 20^2e^{-20}/2!=400 \cdot 0.00000000206/2=0\\\\P(k=3) = 20^3e^{-20}/3!=8000 \cdot 0.00000000206/6=0\\\\P(k=4) = 20^4e^{-20}/4!=160000 \cdot 0.00000000206/24=0\\\\P(k=5) = 20^5e^{-20}/5!=3200000 \cdot 0.00000000206/120=0\\\\P(k=6) = 20^6e^{-20}/6!=64000000 \cdot 0.00000000206/720=0\\\\P(k=7) = 20^7e^{-20}/7!=1280000000 \cdot 0.00000000206/5040=0.001\\\\$

$P(k=8) = 20^8e^{-20}/8!=25600000000 \cdot 0.00000000206/40320=0.001\\\\P(k=9) = 20^9e^{-20}/9!=512000000000 \cdot 0.00000000206/362880=0.003\\\\P(k=10) = 20^{10}e^{-20}/10!=10240000000000 \cdot 0.00000000206/3628800=0.006\\\\P(k=11) = 20^{11}e^{-20}/11!=204800000000000 \cdot 0.00000000206/39916800=0.011\\\\$

$P(k\leq11)=\sum_0^{11} P(k$

b) What is the probability that the number of drivers will exceed 23?

We can write this as:

$P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))$

$P(k=12) = 20^{12}e^{-20}/12!=8442485.238/479001600=0.018\\\\P(k=13) = 20^{13}e^{-20}/13!=168849704.75/6227020800=0.027\\\\P(k=14) = 20^{14}e^{-20}/14!=3376994095.003/87178291200=0.039\\\\P(k=15) = 20^{15}e^{-20}/15!=67539881900.067/1307674368000=0.052\\\\P(k=16) = 20^{16}e^{-20}/16!=1350797638001.33/20922789888000=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=27015952760026.7/355687428096000=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=540319055200533/6402373705728000=0.084\\\\$

$P(k=19) = 20^{19}e^{-20}/19!=10806381104010700/121645100408832000=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=216127622080213000/2432902008176640000=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=4322552441604270000/51090942171709400000=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=86451048832085300000/1.12400072777761E+21=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=1.72902097664171E+21/2.5852016738885E+22=0.067\\\\$

$P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))\\\\P(k>23)=1-(0.021+0.766)=1-0.787=0.213$

c) What is the probability that the number of drivers will be between 11 and 23, inclusive? What is the probability that the number of drivers will be strictly between 11 and 23?

Between 11 and 23 inclusive:

$P(11\leq k\leq23)=P(x\leq23)-P(k\leq11)+P(k=11)\\\\P(11\leq k\leq23)=0.787-0.021+ 0.011=0.777$

Between 11 and 23 exclusive:

$P(11< k$

d) What is the probability that the number of drivers will be within 2 standard deviations of the mean value?

The standard deviation is

$\mu=\lambda =20\\\\\sigma=\sqrt{\lambda}=\sqrt{20}= 4.47$

Then, we have to calculate the probability of between 15 and 25 drivers approximately.

$P(15$

$P(k=16) = 20^{16}e^{-20}/16!=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=0.084\\\\P(k=19) = 20^{19}e^{-20}/19!=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=0.067\\\\P(k=24) = 20^{24}e^{-20}/24!=0.056\\\\$

3 0

3 years ago

Allison had 12 apples in the refrigerator. Her mother went to the grocery store and bought a bag with 24 apples and 12 oranges i

Ksivusya [100]

12 apples+24 apples= 36 apples

3 0

4 years ago

What is an equation of the line that passes through the point (2,-3) and is perpendicular to the line 2x+5y=10?

soldier1979 [14.2K]

Answer:

An equation of the line that passes through the point (2,-3) and is perpendicular to the line is:

$y=\frac{5}{2}x-8$

Step-by-step explanation:

The slope-intercept form of the line equation

$y = mx+b$

where m is the slope and b is the y-intercept

Given the line

$y=-\frac{2}{5}x+2$

Here, the slope = m = -2/5

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:

slope = m = -2/5

perpendicular slope = – 1/m = -1/(-2/5) = 5/2

Using the point-slope form of the line equation

$y-y_1=m\left(x-x_1\right)$

where m is the slope and (x₁, y₁) is the point

substituting the values m = 5/2 and the point (2,-3)

$y-\left(-3\right)=\frac{5}{2}\left(x-2\right)$

$y+3=\frac{5}{2}\left(x-2\right)$

subtract 3 from both sides

$y+3-3=\frac{5}{2}\left(x-2\right)-3$

$y=\frac{5}{2}x-8$

Therefore, an equation of the line that passes through the point (2,-3) and is perpendicular to the line is:

$y=\frac{5}{2}x-8$

8 0

3 years ago

Current research indicates that the distribution of the life expectancies of a certain protozoan is normal with a mean of 46 day

AysviL [449]

Answer: P(x ≥ 47) = 0.25

Step-by-step explanation:

Since the distribution of the life expectancies of a certain protozoan is normal, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = life expectancies of the certain protozoan.

µ = mean

σ = standard deviation

n = number of samples

From the information given,

µ = 46 days

σ = 10.5 days

n = 49

The probability that a simple random sample of 49 protozoa will have a mean life expectancy of 47 or more days is expressed as

P(x ≥ 47) = 1 - P(x < 47)

For x = 47

z = (47 - 46)/(10.5/√49) = 0.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.0.75

P(x ≥ 47) = 1 - 0.75 = 0.25

5 0

3 years ago