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Bas_tet [7]
3 years ago
6

A clothing company produced 600 items of clothing. Of those, 15% were jackets, 25% were dresses, and 15% were shoes. How many of

the items of clothing were NOT jackets, dresses, or shoes?
Mathematics
2 answers:
kenny6666 [7]3 years ago
7 0

Answer:

270 of them were not jackets dresses or shoes

Step-by-step explanation:

Volgvan3 years ago
3 0

Answer:

270 items

Step-by-step explanation:

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The size P of a small herbivore population at time t (in years) obeys the function P(t) = 1000e0-16t if they have enough food an
Stells [14]

Answer:

After 9 years, the population reach 4000.

Step-by-step explanation:

The given function P(t) represents the size of a small herbivore population at time t (in years).

P(t)=1000e^{0.16t}

We need to find the number of years after which population reach 4000.

Substitute P(t)=4000 in the above function.

4000=1000e^{0.16t}

Divide both sides by 1000.

4=e^{0.16t}

Taking ln on both sides.

\ln 4=\ln e^{0.16t}

1.3863=0.16t                 [\because \ln e^x=x]

Divide both sides by 0.16.

\dfrac{1.3863}{0.16}=t

8.664375=t

We need to find the number of years. So, round the answer to the next whole number.

t\approx 9

Therefore, after 9 years, the population reach 4000.

5 0
3 years ago
Mais family drinks a total of 10 gallons of milk every 6 weeks.
Firdavs [7]

Answer:

1 2/3

Step-by-step explanation:

Mai's family drinks a total of 10 gallons of milk every 6 weeks so every week they drink 10/6 = 1 4/6 or 1 2/3 gallons of milk

8 0
3 years ago
In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that
attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable X\sim Poisson(\lambda = 0.62).

The probability function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

             = 1 - P (X = 0) - P (X = 1)

             =1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: \lambda=7\times0.62=4.34

Compute the probability of no deaths in a corps over 7 years as follows:

P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0
3 years ago
What is information given in numbers called? A) facts B)data C) figures D)estimates
yKpoI14uk [10]

Answer:

B) Data

Step-by-step explanation:

Data has to do with numbers.

3 0
4 years ago
Read 2 more answers
Who is correct? Explain your reasoning.
kap26 [50]
In order to get who was right we need to solve the expression:
2a^2b(-2ab^3)^-2
The above can be written as fraction to get:
(2a^2b)/(-2ab^3)^2
=(2a^2b)/(4a^2b^6)
=1/2(a^(2-2)b^(1-6))
=1/2a^0b^-5
=1/2b^(-5)
This implies that neither of the was right

3 0
4 years ago
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