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3 years ago

Leah cut a 7 inch piece of ribbon into pieces that are each 1/2 an inch long.How many pieces of ribbon did she cut?

2 answers:

Ivan3 years ago

8 0

Hey there! :D

If you were to cut 7 pieces in half, then:

1 inch of ribbon= two 1/2 in pieces of ribbon

7*2=14

She cut "C" 14 pieces of ribbon.

I hope this helps!
~kaikers

Sophie [7]3 years ago

3 0

The answer is c) 14 hope this helps

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In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year. Let X be the number of quakes in a given

hichkok12 [17]

Answer:

a) Earthquakes are random and independent events.

b) There is an 85.71% probability of fewer than three quakes.

c) There is a 0.51% probability of more than five quakes.

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 time interval.

In this problem, we have that:

In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year, so $\mu = 1.3$

(a) Justify the use of the Poisson model.

Earthquakes are random and independent events.

You can't predict when a earthquake is going to happen, or how many are going to have in a year. It is an estimative.

(b) What is the probability of fewer than three quakes?

This is $P(X < 3)$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725$

$P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543$

$P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2725 + 0.3543 + 0.2303 = 0.8571$

There is an 85.71% probability of fewer than three quakes.

(c) What is the probability of more than five quakes?

This is $P(X > 5)$

We know that either there are 5 or less earthquakes, or there are more than 5 earthquakes. The sum of the probabilities of these events is decimal 1.

$P(X \leq 5) + P(X > 5) = 1$

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725$

$P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543$

$P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303$

$P(X = 3) = \frac{e^{-1.3}*1.3^{3}}{(3)!} = 0.0980$

$P(X = 4) = \frac{e^{-1.3}*1.3^{4}}{(4)!} = 0.0324$

$P(X = 5) = \frac{e^{-1.3}*1.3^{5}}{(5)!} = 0.0084$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2725 + 0.3543 + 0.2303 + 0.0980 + 0.0324 + 0.0084 = 0.9949$

Finally

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9949 = 0.0051$

There is a 0.51% probability of more than five quakes.

5 0

3 years ago

Which number is the additive inverse of -10 1/4 A 4/41. B -4 1/10 C 0 D 10 1/4

Juli2301 [7.4K]

Answer = 10 1/4 (Letter D)

3 0

2 years ago

Read 2 more answers

The Mars Cereal Company has two different cereal boxes for Mars Cereal. The large box is 8 inches wide, 11 inches high, and 3 in

lubasha [3.4K]

Answer:

a)90 square inches

b)104 cubic inches

Step-by-step explanation:

Dimensions of the bigger box

W= 8 in

H=11 in

D= 3 in

Dimensions of the smaller box

w= 6 in

h= 10 in

h= 2.5 in

therefore, difference in the area of the cupboard

=2[ WH+HD+WD] - 2[wh+hd+wd]

= 2[8*11+11*3+8*3]- 2[6*10+10*2.5+6*2.5]

=290- 200= 90 square inches

volume of the large cupboard = WHD= 8*11*3= 264 cubic in

volume of the small cupboard= whd= 6*10*2.5= 150 cubic in

therefore, 264-150= 114 cubic in more cereal does the large box hold than the small box

6 0

3 years ago

Read 2 more answers

My answer is 168m Area of triangle. Base = 16m, height 21m.

Karolina [17]

Area = 16 × 21 × 1/2
= 168 m

your answer was right !!! keep on practice

7 0

3 years ago

Which graph do I use on number 3-5 HELP ASAP PLEASE!!!

Kisachek [45]

Answer:

[5, -8] + [0, -5] + [-7, 0] + [0, -6] = [-2, -19] --> III. Quadrant

[-2, 6] + [0, 3] + [5, 0] + [0, 4] = [3, 13] --> I. Quadrant

[0, -5] + [-5, 0] + [0, 4] + [7, 0] + [0, -3] = [2, -4] --> IV. Quadrant

8 0

3 years ago