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Papessa [141]
3 years ago
13

Simplify 28 over 35

Mathematics
2 answers:
maria [59]3 years ago
8 0
4 / 5
use the 7 to reduce each numer of the fraction
bearhunter [10]3 years ago
8 0
If we simplify 28/35 it will equal 4/5. You use the Greatest Common Factor to divide it which is 7 and it will equal 4/5
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Jim had some dimes, quarters, and nickels. First, he counted only the dimes and quarters and found that he had 9 coins. Then, he
Olegator [25]

Answer:

$1.90

Step-by-step explanation:

Represent the numbers of nickels, dimes and quarters by n, d and q.

Then: d + q = 9, or q = 9 - d.

Also, n + d = 10

Lastly, n + d + q = 15.

Let's substitute 9 - d for q in the equation immediately above:

n + d + (9 - d) = 15, or n + 9 = 15.  Then n must be 6.

In summary, Jim has 6 nickels, (10 - 6) dimes and (9 - 4) quarters, or:

                                    6 nickels, 4 dimes and 5 quarters

Thus, he has 5($0.05) + 4($0.10) + 5($0.25), or

                         $0.25   +   $0.40   + $1.25, or          $1.90

7 0
3 years ago
Find the value of n that makes ΔDEF ∼ΔXYZ when DE = 4, EF = 5, XY = 4(n+1), YZ = 7n - 1, and ∠E ≅∠Y. n =
dalvyx [7]

Answer:

3

Step-by-step explanation:

It's given that ΔDEF ∼ΔXYZ . So the corresponding sides of both triangles will be proportional to each other.

=  > \frac{de}{xy}  =    \frac{ef}{yz}  =  \frac{df}{xz}

DE = 4 ; XY = 4(n + 1) ; EF = 5 ; YZ = 7n - 1

Putting all these values gives ,

\frac{4}{4(n + 1)}  =  \frac{5}{7n - 1}

=  >  \frac{1}{n + 1}  =  \frac{5}{7n - 1}

=  > 7n - 1 = 5(n + 1)

=  > 7n - 1 = 5n + 5

=  > 7n - 5n = 5 + 1

=  > 2n = 6

=  > n =  \frac{6}{2}  = 3

3 0
3 years ago
Read 2 more answers
How do you estimate21× 78
finlep [7]

21 *78=

20 * 80=

1600 as an estimate


8 0
3 years ago
The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x
RoseWind [281]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

       P(a < X < b) =  \int_{a}^{b} \frac{2}{x^{3}} dx = 2\int_{a}^{b} x^{-3} dx

                            = 2[ \frac{x^{-3+1} }{-3+1}]^{b}_a   dx    { Because \int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a }

                            = 2[ \frac{x^{-2} }{-2}]^{b}_a = \frac{2}{-2} [ x^{-2} ]^{b}_a

                            = -1 [ b^{-2} - a^{-2}  ] = \frac{1}{a^{2} } - \frac{1}{b^{2} }

a) Now P(X < 5) = P(1 < X < 5)  {because x > 1 }

     Comparing with general probability we get,

     P(1 < X < 5) = \frac{1}{1^{2} } - \frac{1}{5^{2} } = 1 - \frac{1}{25} = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/8^{2} - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = \frac{1}{6^{2} } - \frac{1}{10^{2} } = \frac{1}{36} - \frac{1}{100 } = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

                                = (\frac{1}{1^{2} } - \frac{1}{6^{2} }) + (1/10^{2} - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

               ⇒  P(1 < X < x) = 0.75

               ⇒  \frac{1}{1^{2} } - \frac{1}{x^{2} } = 0.75

               ⇒  \frac{1} {x^{2} } = 1 - 0.75 = 0.25

               ⇒  x^{2} = \frac{1}{0.25}   ⇒ x^{2} = 4 ⇒ x = 2  

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0
3 years ago
M/18=24/15 <br>how do I do this problem?
GrogVix [38]
M/18 = 24/15
15m = 18 * 24
15m = 432
m = 28.8
4 0
3 years ago
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