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3 years ago

Hi I need help with this problem. But it has to be shown with work. Does anyone know?

Mathematics

2 answers:

SpyIntel [72]3 years ago

8 0

Answer:

0.4 or 2/5???

Step-by-step explanation:

I believe this is a proportion/ratio problem.

Think of it like this:

x/6 = 10/150

To solve for x, you must cross multiply:

150x = 6*10 (150x... DON'T write 150*x, that is nonsense)

150x = 60

x = 60/150 (you divide because you are actually dividing 150 on both sides and doing the opposite operation of multiplication)

x = 6/15

x = 2/5 or 0.4 (I simplified it)

Hopefully this helped? I can't seem to find a whole number answer... so the only options I could do are 2/5 or 0.4

Damm [24]3 years ago

5 0

Answer:

its is 90

Step-by-step explanation:

6x10=60 150-60=90

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What is one third the difference between 3/4 and 2/3?

Ivan

Answer:

One third the difference between 3/4 and 2/3 is $(\frac{1}{36} )$

Step-by-step explanation:

One third the difference of A and B = $\frac{1}{3} \times (A-B)$

Here, as given in the question:

$A = \frac{3}{4} , B = \frac{2}{3} \\\implies A - B = \frac{3}{4} - \frac{2}{3}\\=(\frac{3 \times 3 -2 \times 4}{12}) = (\frac{9-8}{12} ) = (\frac{1}{12})\\ \implies A - B = (\frac{1}{12})$

Now, one third of the difference is :

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Hence, one third the difference between 3/4 and 2/3 is $(\frac{1}{36} )$

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4 years ago

What are the equivalent fractions with a common denominator for 2/5 and 1/2?

Svetlanka [38]

So 1/2 is 5/10 and 2/5 is 4/10 common denonominator is 10 ( i think )

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3 years ago

Read 2 more answers

the diameters of Douglas firs grown at a Christmas tree farm are normally distributed with a mean of 4 inches and a standard dev

aksik [14]

Answer:

Proportion of the trees will have diameters between 2 and 6 inches = 0.8164

Step-by-step explanation:

Given -

Mean $(\nu )$ = 4

Standard deviation $(\sigma )$ = 1.5

Let X be the diameter of tree

proportion of the trees will have diameters between 2 and 6 inches =

$P(2< X< 6)$ = $P(\frac{2 - 4 }{1.5}< \frac{X - \nu }{\sigma}< \frac{6 - 4 }{1.5})$

= $P(\frac{-2 }{1.5}< Z< \frac{2 }{1.5})$ Put [ $Z = \frac{X - \nu }{\sigma}$ ]

= $P(-1.33< Z< 1.33)$

= $(Z< 1.33) - (Z< -1.33)$

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