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allsm [11]
3 years ago
11

28/48 in the simplest form

Mathematics
2 answers:
yulyashka [42]3 years ago
4 0
I agree its 7/12....
mr Goodwill [35]3 years ago
3 0
Expressed as a proper fraction in its simplest form, by dividing the numerator and denominator by 4, 28/48 is equal to 7/12 or seven twelfths.
You might be interested in
6r - 7 over 10 = r over 4
Ahat [919]
Multiply both sides by 4

subtract 24r from both sides

simplify r - 24r to -23r

divide both sides by -23

two negatives make a positive

simplify 14/5/23 to 14/5 x 23

simplify 5 x 23 to 115

switch sides

Answer: r = 14/115.
7 0
3 years ago
a store owner wishes to make a new tea with a unique flavor by mixing black tea and oolong tea. if he has 35 pounds of oolong te
Verizon [17]

35 pounds of black tea is needed to be mixed to get the final mixture for 2.10 per pound

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the amount of black tea worth 1.80 per pound to be mixed to get the final mixture for 2.10 per pound, hence:

2.4(35) + 1.8(x) = 2.1(x + 35)

x = 35 pounds

35 pounds of black tea is needed to be mixed to get the final mixture for 2.10 per pound

Find out more on equation at: brainly.com/question/2972832

#SPJ1

3 0
1 year ago
What is the value of 2/3^-4
Reptile [31]

Answer:

\frac{16}{49}

Step-by-step explanation:

To solve problems like this, we need to multiply the base ,\frac{2}{3} , the amount of times as the exponent, 4.

Essentially, the equation is \frac{2}{3} x \frac{2}{3}  x \frac{2}{3}  x \frac{2}{3} .

The product would be \frac{16}{49}.

Now we can't forget that the exponent is a negative. But because the exponent is an even number, we don't need to worry about that.

Hope this helped :)

8 0
3 years ago
Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

3 0
2 years ago
Find the limit
Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

\large\underline{\sf{Solution-}}

Given expression is

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

On substituting directly x = 1, we get,

\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}

\rm \: = \sf \: \: - \infty \: - \: \infty

which is indeterminant form.

Consider again,

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

can be rewritten as

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]

\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}

\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}

\rm \: = \: \sf \boxed{2}

Hence,

\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}

\rule{190pt}{2pt}

7 0
2 years ago
Read 2 more answers
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