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3 years ago

9 Check: 6,350 - 2,460

Mathematics

1 answer:

bazaltina [42]3 years ago

5 0

Answer:

3890

Step-by-step explanation:

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HELP!!!!!! An Instagram photo is measured ″ 4″ by ″ 4″. A poster enlargement is being printed with a scale factor of 8.5. What i

Anika [276]

The given measurement of the Instagram photo is "4" by "4". And a poster enlargement is being printed with a scale factor of 8.5 .

So on using the scale factor of 8.5 , lengths of new sides are "8.5 *4" by "8.5 *4"

= " 34" by "34" .

So the new lengths of the enlarged pictures are of dimension "34" by "34" .

7 0

3 years ago

Read 2 more answers

A rectangle's length is twice its width. if the perimeter is 132 cm, find the width of the rectangle.

Over [174]

Perimeter=(2*length)+(2*width)
length=2*w
132=(2(2w))+(2w)
132=(4w)+(2w)
132=6w
w=22
l=44

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3 years ago

Explain to me how to do this please

nataly862011 [7]

You must first attach the problems with your question.

6 0

3 years ago

Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b

jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .

So; as each place is fitted with either 0 or 1

$\mathbf{|R|= 2^{10}}$

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

$\mathbf{P(E) = \dfrac{|E|}{|R|}}$

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:

$\mathbf{{|E|}=1 }$ ; $\mathbf{|R|= 2^{10}}$

$\mathbf{P(E) = \dfrac{1}{2^{10}}}$

4 0

3 years ago

Please help me with this math question

scZoUnD [109]

Answer:

Step-by-step explanation:

$\bold{METHOD\ 1}\\\\6d=8e\qquad\text{divide both sides by 6}\\\\d=\dfrac{8}{6}e\to d=\dfrac{4}{3}e\\\\\text{Substitute it and}\ f=2e\ \text{to the ratio}\ d:e:f:\\\\\dfrac{4}{3}e:e:2e\qquad\text{cancel}\ e\\\\\dfrac{4}{3}:1:2\qquad\text{multiply by 3}\\\\\huge\boxed{4:3:6}$

$\bold{METHOD\ 2}\\\\6d=8e\qquad\text{divide both sides by 6}\\\\d=\dfrac{8}{6}e\qquad\text{divide both sides by}\ e\neq0\\\\\dfrac{d}{e}=\dfrac{4}{3}\to d:e=4:3\\\\2e=f\qquad\text{divide both sides by 2}\\\\e=\dfrac{f}{2}\to e=\dfrac{1}{2}f\qquad\text{divide both sides by}\ f\neq0\\\\\dfrac{e}{f}=\dfrac{1}{2}\to e:f=1:2\\\\\text{We have}\ d:e=4:3\ \text{and}\ e:f=1:2.$

$\text{We know}\ \dfrac{1}{2}=\dfrac{1\cdot3}{2\cdot3}=\dfrac{3}{6}\\\\\text{threfore}\ e:f=3:6\\\\\text{Now we have}\\\\\left\begin{array}{ccc}d:e=4:3\\e:f=3:6\end{array}\right\}\large{\boxed{d:e:f=4:3:6}}$

6 0

3 years ago