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4 years ago

Hey if you could help that be great I've stuck on this for awhile

Mathematics

1 answer:

n200080 [17]4 years ago

5 0

Conversions aren't bad until you have to deal with the cubed and squared versions of them!! Remember that there are 100 cm in 1 meter. If that be the case, then there are 100cm*100cm*100cm (100cm^3) in m*m*m (1 meter cubed). We will use that in our conversion then: $250,000 \frac{cm^3}{m^2}* \frac{m*m*m}{100cm*100cm*100cm}$ . The cm^3 cancels out leaving m^3 over m^2 which is what we want. Doing the math now we get $.25 \frac{m^3}{m^2}$ , or $\frac{1}{4} \frac{m^3}{m^2}$

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The fractions with the same numerators and different denominators are NOT equivalent.

WINSTONCH [101]

B has the same denominator so there is your answer

7 0

3 years ago

Help me please and i give you points

KatRina [158]

Answer:

30-100

28-100

25-100

7-100

Step-by-step explanation:

It's hard to explain but I try

part in this means how many seeds we can place in those parts

and the whole it he whole amount of the part

hope that make sense

8 0

3 years ago

For what power of q is its value 1?

DedPeter [7]

Answer:

0, for q ≠ 0 and q ≠ 1

Step-by-step explanation:

Assuming q ≠ 0, you want to find the value of x such that ...

q^x = 1

This is solved using logarithms.

x·log(q) = log(1) = 0

The zero product rule tells us this will have two solutions:

x = 0

log(q) = 0 ⇒ q = 1

If q is not 0 or 1, then its value is 1 when raised to the 0 power. If q is 1, then its value will be 1 when raised to any power.

_____

Additional comment

The applicable rule of logarithms is ...

log(a^b) = b·log(a)

4 0

2 years ago

Write an expression that represents the width of a rectangle with length x+5 and area x^3+12x^2+47x+60

xeze [42]

Area divided by length= width
x^3+12x^2+47x+60 ÷ x+5=w

6 0

3 years ago

A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occ

Shkiper50 [21]

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

$\lambda=r*t=0.000011*13,280=0.14608$

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

$P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990$

c) The probability of more than one case is:

$P(k>1)=1-P(k\leq 1)=1-0.990=0.010$

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

7 0

3 years ago