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3 years ago

A girl walks 5000 meters in 1 hour if the girl walks the same speed for 4 hours how many kilometers would you have walked

2 answers:

5 0

She would have walked 20 km because if you multiply 5000m by 4(hours) you get 20,000m. And there are 1000 m in one km, so take 20,000 and divide it by 1000, therefore you get 20 km

Hatshy [7]3 years ago

4 0

The girl will have walked 20,000 meters or 20 km

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A candy distributor needs to mix 10% a fat content chocolate with 40% fat content chocolate to create 200 kg of a 16% fat conten

Luden [163]

I like to use an X diagram to solve problems like this. The constituent percentages are on the left, the desired mix is in the middle and the numbers on the diagonals are the difference between the mix and the constituent.

The numbers 6 and 24 here represent the ratio of 40% to 10% chocolate. That 6:24 ratio can be reduced to 1:4. The sum of those ratio units is 1+4 = 5, so each must represent (200 kg)/5 = 40 kg of chocolate.

The mix must consist of
40 kg of 40% fat chocolate
160 kg of 10% fat chocolate

6 0

3 years ago

Let f(x) = -4x + 7 and g(x) = 10x - 6 find f(g(x))

max2010maxim [7]

F(x) = -4x + 7
g(x) = 10x - 6

f(g(x)) = -4(10x - 6) + 7
f(g(x)) = -4(10x) + 4(6) + 7
f(g(x)) = -40x + 24 + 7
f(g(x)) = -40x + 31

7 0

3 years ago

Read 2 more answers

Triangle ABC has vertices at A(-1,1), B(-7,1), and C(-3,6).<br> What is the area of Triangle ABC?

AURORKA [14]

Answer:

I think the answer is b

Step-by-step explanation:

6 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

Simplify the equation

Arturiano [62]

Answer:

$\dfrac{16x^3-37x^2-16x+37}{x^2+2}$

Step-by-step explanation:

There aren't any factors that cancel (except 3x). The best you can do is multiply it out.

$\displaystyle\frac{\frac{3x^2-3}{x}}{\frac{3(x^2+2)}{16x^2-37x}}=\frac{3(x^2-1)}{x}\cdot\frac{16x^2-37x}{3(x^2+2)}\\\\=\frac{x(x^2-1)(16x-37)}{x(x^2+2)}=\frac{16x^3-37x^2-16x+37}{x^2+2}$

7 0

3 years ago