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4 years ago

Which expressions are equivalent to -3 times 4/-5?

1 answer:

8 0

Hi, I commented on your previous question but never got a response :(

-3 times 4 divided by -5 = -3x-4 is 12, 12/-5 is -2.4
-3 times -4/5 is 12, 12/5 is 2.4

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Write the following ratios in its simplest form 56:112:28

damaskus [11]

Answer:

2 : 4 : 1

Step-by-step explanation:

First take common factors by factorization.

56 = 14 x 4 = 28 x 2

112 = 14 x 8 = 28 x 4

28 = 14 x 2 = 28 x 1

Remove the 28 common factor.

Therefore,

56 : 112 : 28

= 2 : 4 : 1

5 0

3 years ago

Which list shows fraction that ALWAYS result in a terminating decimal for values of n > 0? A) 1/11, 1/121, 1/1331 . . . , 1/1

Alla [95]

Answer:

B) 1/10, 1/100, 1/1000 . . . , 1/10n

Step-by-step explanation:

Comparing all the options given:

A) 1/11, 1/121, 1/1331 . . . , 1/11n

Finding the decimal of the first term,

1/11 = 0.0909090 (never ending decimal place)

B) 1/10, 1/100, 1/1000 . . . , 1/10n

Finding the decimal of the first term,

1/10 = 0.01 (ended decimal place)

C) 1/6, 1/36, 1/126 . . . , 1/6n

Finding the decimal of the first term,

1/6 = 0.16666667 (never ending decimal place)

D) 1/3, 1/9, 1/27 . . . , 1/3n

Finding the decimal of the first term,

1/3 = 0.33333333 (never ending decimal place)

7 0

3 years ago

Read 2 more answers

For Elizabeth's lemonade recipe, 4 lemons are required to make 2 cups of lemonade. At what rate are lemons being used in cups of

Scorpion4ik [409]

Answer:

2 lemons per cup

Step-by-step explanation:

4/2 is two.

6 0

3 years ago

Read 2 more answers

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

HELP PLS ASAP:) What is the slope of the line plotted below? A. -1 B. 1 C. -0.5 D. 0.5

EleoNora [17]

You can solve the by dividing the change in y by the change in x. the answer should be D

7 0

3 years ago

Read 2 more answers