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Vlad1618 [11]
3 years ago
12

The following table shows the total number of quizzes taken in high school. How many quizzes are taken per week?

Mathematics
2 answers:
Anastaziya [24]3 years ago
8 0

thx i needed this answer i have 100 Q's test *sigh*

igomit [66]3 years ago
3 0

Answer:

4

Step-by-step explanation:

y÷x

12÷3=4 and 20÷5=4

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Aliens have injected Ana with mathematical nanobots. These nanobots force Ana to think about math more and more. Initially, ther
natulia [17]

Answer:

<em>Math will take over Ana's brain at 4.4 hours</em>

Step-by-step explanation:

<u>Exponential Grow </u>

The population of the nanobots follows the equation  

p(t) = 5\cdot 2^t

We must find the value of t such that the population of nanobots is 106 or more, that is

5\cdot 2^t\geq 106

We'll solve the equation

5\cdot2^t= 106

Dividing by 5

2^t= 106/5=21.2

Taking logarithms

log(2^t)= log(21.2)

By logarithms property

t\cdot log(2)= log(21.2)

Solving for t

\displaystyle t=\frac{log21.2} {log2}

t=4.4 \ hours

Math will take over Ana's brain at 4.4 hours

6 0
3 years ago
I got the degree to be 81.81818182 I just don't know how get the minute please do tell how to solve by using calculator or by pa
lutik1710 [3]
Well a clock has twelve hours that make up a full circle, or 360 degrees, so you divide 12 by 360 then multiply by the degree, which in this case is 81.81 and that's the number of hours. (You figure out the minutes by looking at the decimal)
6 0
3 years ago
A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 6%
Georgia [21]

Answer:

a

\% E =  0.9 \%

b

\%E_1 = 10.1 \%

Step-by-step explanation:

From the question we are told that

The probability that an employees suffered lost-time accidents last year is P(e) =  0.06

The probability that an employees suffered lost-time accident during the current year is

P(c) =  0.05

The probability that an employee will suffer lost time during the current year given that the employee suffered lost time last year is

P(c | e) =  0.15

Generally the probability that an employee will experience lost time in both year is mathematically represented as

P(c \ n \ e) =  P(e) *  P(c \ |\ e)

=> P(c \ n \ e) =  0.06*   0.15

=> P(c \ n \ e) = 0.009

Generally the percentage of employees that will experience lost time in both year is mathematically represented as

\% E =  P(e \ n \ c ) * 100

=> \% E =  0.009 * 100

=> \% E =  0.9 \%

Generally the probability that an employee will experience at least one lost time accident over the two-year period is mathematically represented as

P(e \ u \ c) =  P(e) + P(c) - P(e \ n \  c)

=> P(e \ u \ c) =  0.06 + 0.05 - 0.009

=> P(e \ u \ c) =  0.101

Generally the percentage of the employees who will suffer at least one lost-time accident over the two-year period is mathematically represented as

\%E_1 = P(e \ u \ c) *  100

=> \%E_1 = 0.101*  100

=> \%E_1 = 10.1 \%

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