All categories

4 years ago

If you blink your eyes 65 times in 5 minutes, how many times would you blink your eyes in 80 minutes?

Mathematics

2 answers:

ivann1987 [24]4 years ago

5 0

Answer:

1,040

Step-by-step explanation:

65/5 = x/80

5x = 5200

/5 /5

x=1,040

Vikentia [17]4 years ago

3 0

5,200 is the answer

You might be interested in

The data recresent the results for a test fx a certain discesa. Assume che ritual from the group is ratasty sieded. Fedtreprobab

tigry1 [53]

The total sum of students given four quarters = The student who spent the money on gum plus the students that kept the money.

$34+17=51$

The total sum of students given four quarters is 51 students.

The student who spent the money on gum is 34 students.

The probability of the students that spent the money, given that the students were given four quarters,

$\frac{\text{The probability of students that spent the money}}{\text{The total sum of students given four quarters}}=\frac{34}{51}$

The probability is,

$\frac{34}{51}=\frac{2}{3}=0.66667\approx0.667(3\text{ decimal places)}$

The probability is 0.667.

8 0

1 year ago

7. Three cubes of edges x cm , 6 cm, and 8 cm are melted and recast into a bigger cube of

Llana [10]

Answer:

10 cm³

Step-by-step explanation:

Volume of a cube = (Side)³

Volume of bigger cube = (12)³

(12)³ = 1728 cm³

Volume of three cubes = (x)³ + (6)³ + (8)³ = 1728

(x)³ = 1728 - (6)³ - (8)³

(x)³ = 1000

x = ∛1000

x = 10 cm³

8 0

3 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

What type of triangle is defined by the following set of side lengths

NARA [144]

Answer:

ACUTE TRIANGLE 

Step-by-step explanation:

If the sum of the squares of the two shorter sides of a triangle is greater than the square of the longest side, the triangle is acute. However, if the sum of the squares of the two shorter sides of a triangle is smaller than the square of the longest side, the triangle is obtuse.

Therefore, the triangle with sides 11, 9, and 7 is an acute triangle since 9² + 7² is greater than 11²

${9}^{2} \: + \: {7}^{2} > \: {11}^{2} \\ 130 > 121$

8 0

3 years ago

Define mean deviation and standard deviation.

Alina [70]

Answer:

The standard deviation is simply the square root of the variance. The average deviation, also called the mean absolute deviation, is another measure of variability. To calculate the average deviation, simply subtract the mean from each value, then sum and average the absolute values of the differences.

3 0

3 years ago