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2 years ago

Lillie jogs 1/2 mile in 1/10 hour what is Lilli's rate in miles per hour

Mathematics

2 answers:

Goshia [24]2 years ago

6 0

U technecally wrote the answer

RoseWind [281]2 years ago

4 0

The rate in miles per hour Lillie jogs is 5 miles per hour.

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hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams a

nalin [4]

Answer:

a. 205320

b. 34220

c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r)!

n= number of things to choose from

r= Choosing r number

60P3= 60! / (60 - 3)!

(60)(59)(58)(57)! / (57)!

=205320

B) The number of ways to dustribute the exams among the TA's is:

n! /(n - r)! r!

60C3= 60! /(60 - 3)! 3!

= 60!/ 57! 3!

= 60 × 59 × 58 / 3 × 2 × 1

= 34220

C) The required number of ways is:

60C25 + 60C20 + 60C15

= 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

6 0

2 years ago

The cost of insurance on a house is directly proportional to the area of the house.Samhas a 2,500 square-foot house and pays a m

DochEvi [55]

Answer:

y tú la vez bebiendo de la botella guz dividido por qué colgaron que no se sienta en tu casa no te preocupes que no me dijan no me dijan no me dijan que hacen

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2 years ago

Which of the following is the graph of y = negative StartRoot x EndRoot + 1? On a coordinate plane, an absolute value curve curv

mestny [16]

Answer:

B. On a coordinate plane, an absolute value curve curves up and to the right in quadrant 4 and starts at y = 1.

Step-by-step explanation:

<u>Graph of the function:</u>

y = -√x + 1

The domain is x ≥ 0, the range y ≤ 1

Correct answer choice is B

On a coordinate plane, an absolute value curve curves up and to the right in quadrant 4 and starts at y = 1.

<em>The graph is attached</em>

7 0

2 years ago

Read 2 more answers

let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as expo

anygoal [31]

Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

$f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\$

a) P(x > 5) = $\int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865$

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = $\int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}$

But P(x > a) = 0.05

Therefore:

$e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98$

a ≅ 12

6 0

2 years ago

Value of expression 6 3/4(-11.5)

Olenka [21]

Answer:

6 * 3/4 = 9 2 = 412 = 4.5

Step-by-ste

6 * 3/4 = 6/1 3*4 = 18/4 = 9/2

2 · 2 = 9/2

Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(18, 4) = 2. In the next intermediate step cancelling by a common factor of 2 gives 9

In words - six multiplied by three quarters = nine halfs.

0 0

2 years ago