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3 years ago

Maggie has three pages of homework to do.If she can do 1/4 of a page in one hour, how many will her homework take

Mathematics

2 answers:

Bad White [126]3 years ago

6 0

Answer:

the answer is 12 :)))))

hodyreva [135]3 years ago

3 0

Answer:

The answer is 12 hours.

Step-by-step explanation:

If she completes 1/4 of a page in an hour completing one page will take her 4 hours after that you multiply it by 3 and get the total of 12 hours that is takes for her to do her homework

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lesantik [10]

Answer:

Step-by-step explanation:

To find the mean;

add up the data

(50+60+60+70) =240

Then divide by the number of data points

240/4 = 60

The mean is 60

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3 years ago

How much string is out if a kite is 100 feet above the ground and the string makes an angle of 65° with the ground? Draw a diagr

aleksklad [387]

Answer:

I believe its is 90.6 for plato if i remember correctly

Step-by-step explanation:

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4 years ago

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Solve 7x-2x(x+1)=6x+14 A) x=-12 B) x=12 C) x=-16 D) x=16

prisoha [69]

Answer:

x=-16 (answer : C

Step-by-step explanation:

hello:

put x=-16 in this equation : 7x-2(x+1)=6x+14

7(-16)-2(-16+1)=6(-16)+14

-112-2(-15) = -96+14

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-82 = -82 .....true

3 0

3 years ago

The average number of annual trips per family to amusement parks in the UnitedStates is Poisson distributed, with a mean of 0.6

IrinaK [193]

Answer:

a) 0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b) 0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c) 0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d) 0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e) 0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Poisson distributed, with a mean of 0.6 trips per year

This means that $\mu = 0.6n$ , in which n is the number of years.

a.The family did not make a trip to an amusement park last year.

This is P(X = 0) when n = 1, so $\mu = 0.6$ .

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0)!} = 0.5488$

0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b.The family took exactly one trip to an amusement park last year.

This is P(X = 1) when n = 1, so $\mu = 0.6$ .

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293$

0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c.The family took two or more trips to amusement parks last year.

Either the family took less than two trips, or it took two or more trips. So

$P(X < 2) + P(X \geq 2) = 1$

We want

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.5488 + 0.3293 = 0.8781$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8781 = 0.1219$

0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d.The family took three or fewer trips to amusement parks over a three-year period.

Three years, so $\mu = 0.6(3) = 1.8$ .

This is

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.8}*(1.8)^{0}}{(0)!} = 0.1653$

$P(X = 1) = \frac{e^{-1.8}*(1.8)^{1}}{(1)!} = 0.2975$

$P(X = 2) = \frac{e^{-1.8}*(1.8)^{2}}{(2)!} = 0.2678$

$P(X = 3) = \frac{e^{-1.8}*(1.8)^{3}}{(3)!} = 0.1607$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1653 + 0.2975 + 0.2678 + 0.1607 = 0.8913$

0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e.The family took exactly four trips to amusement parks during a six-year period.

Six years, so $\mu = 0.6(6) = 3.6$ .

This is P(X = 4). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.1912$

0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

4 0

3 years ago

H E L P M E P L E A S E !!!

Art [367]

Given:

The table of data values.

To find:

What is the percentage of 10th grade students are in 2 clubs.

Solution:

From the given table it is clear that the number of 10th grade students are in 2 clubs is 12.

Total number of students in 10th grade is:

$125+12+15=152$

Now, the percentage of 10th grade students are in 2 clubs is:

$\text{Required percentage}=\dfrac{\text{Number of 10th grade students are in 2 clubs}}{\text{Total number of students in 10th grade}}\times 100$

$\text{Required percentage}=\dfrac{12}{152}\times 100$

$\text{Required percentage}=\dfrac{1200}{152}$

$\text{Required percentage}\approx 7.89\%$

Therefore, the percentage of 10th grade students are in 2 clubs is 7.89%.

5 0

3 years ago