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dolphi86 [110]
4 years ago
5

Find the qoutient of 1170 and 45

Mathematics
2 answers:
statuscvo [17]4 years ago
7 0

Answer:

36

Step-by-step explanation:

MissTica4 years ago
5 0
Quotient is another word for divide
1170/45
the answer is 26
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Do not round intermediate calculations. round your answers to two decimal places. a. define a random variable number of times th
navik [9.2K]

Observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units 1.

Total no. of times owner-occupied units had a water supply stoppage lasting 6 or more hours

547000 + 5012000 + 6110000 + 2544000 + 557000 = 14770000

x = no. of times owner-occupied units had a water supply stoppage.

<h3>What is the probability at x =0?</h3>

P(x=0) = 547000/14770000

P(x=0) = 0.0370

Similarly, we have at x=1

P(x=1) = 5012000/14770000

P(x=1) = 0.3393

P(x=2) = 6110000/14770000

P(x=2) = 0.4137

P(x=3) = 2544000/14770000

P(x=3) = 0.1722

P(x=4) = 557000/14770000

P(x=4) = 0.0378

x           f(x)

0             0.0370

1              0.3393

2         0.4137

3              0.1722

4         0.0378

Total     1

Observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units 1.

To learn more about the probability distribution visit:

brainly.com/question/24756209

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An equation is shown 120÷ __ =1.20
devlian [24]

an equation is shown 120÷ __ =1.20

120 / x = 1.2

120/100 = 1.2

check:

1.2* 100 = 120

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3 years ago
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How do I solve 6v-5/8=7/8
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6v-5/8=7/8
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3 years ago
Uninhibited growth can be modeled by exponential functions other than​ A(t) ​=Upper A 0 e Superscript kt. For ​ example, if an i
laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than A(t)=A_{0}e^{kt}. for example, if an initial population P₀ requires n units of time to triple, then the function P(t)=P_{0}(3)^{\frac{t}{n} } models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form P(t)=P_{0}(3)^{\frac{t}{n} } that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form A(t)=A_{0}e^{kt}.

Answer: a) P(t)=50(3)^{\frac{t}{30} }

              b) P(t) = 280 insects

              c) t = 74 days

             d) A(t)=50e^{0.037t}

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

P(t)=50(3)^{\frac{t}{30} }

b) In t = 47 days:

P(t)=50(3)^{\frac{t}{30} }

P(47)=50(3)^{\frac{47}{30} }

P(47)=50(3)^{1.567}

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

750=50(3)^{\frac{t}{30} }

\frac{750}{50}=(3)^{\frac{t}{30} }

(3)^{\frac{t}{n} }=15

Using the property <u>Power</u> <u>Rule</u> of logarithm:

log(3)^{\frac{t}{30} }=log15

\frac{t}{30}log(3)=log15

t=\frac{log15}{log3} .30

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

A(t)=A_{0}e^{kt}

3A_{0}=A_{0}e^{30k}

3=e^{30k}

Use Power Rule again:

ln3=ln(e^{30k})

ln3=30k

k=\frac{ln3}{30}

k = 0.037

Equation for exponential growth will be:

A(t)=50e^{0.037t}

3 0
3 years ago
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