All categories

3 years ago

2/15 voted omar, 3/5 voted scott, how many voted for Lindsay?

Mathematics

1 answer:

Talja [164]3 years ago

7 0

2/15 and 3/5 share 15 as the least common denominator, hence, added up, it equals 11/15. Lindsay has 4/15 votes.

You might be interested in

2 5/8 times 1 1/7 equals

Alex17521 [72]

168/56 or 3. 21/8 times 8/7 equals 168 over 56. Also can be 3

6 0

3 years ago

Read 2 more answers

a line passes through point (-2,5) and has a slope of 2/3 points a (x,3) and b ( -2, y) lie on the line what is the value of y a

Tomtit [17]

Answer:

The value of y is 5 , and The value of x is - 5

Step-by-step explanation:

Given as :

The line passes through point ( - 2, 5)

The slope of line is m = $\frac{2}{3}$

Equation of line is written as

y - $y_1$ = m × (x - $x_1$ )

Since line passes through line ( - 2 , 5 ) and have slope $\frac{2}{3}$

So, y - 5 = $\frac{2}{3}$ × (x + 2)

Or, 3 × ( y - 5 ) = 2 × ( x + 2 )

Or, 3 y - 15 = 2 x + 4

Or, 3 y - 2 x - 19 = 0 is the equation of line

Now points ( x , 3 ) lies on the line 3 y - 2 x - 19 = 0

So, 3 × 3 - 2 x - 19 = 0

or, 9 - 2 x - 19 = 0

or, 2 x = - 10

∴ x = - $\frac{10}{2}$

I.e x = - 5

Again points ( - 2 , y ) lies on the line 3 y - 2 x - 19 = 0

So, 3 × y + 2 × 2 - 19 = 0

Or, 3 × y + 4 -19 = 0

or, 3 y = 15

∴ y = $\frac{15}{3}$

I.e y = 5

Hence The value of y is 5 , and The value of x is - 5 . Answer

8 0

3 years ago

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi

Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}$

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

$P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483$

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

5 0

3 years ago

Can someone please answer. There is only one problem. There is a picture. Thank you!!!

Rainbow [258]

The number of permutations of the 25 letters taken 2 at a time (with repetitions) is: $25^{2}$
The number of permutations of the 9 digits taken 4 at a time (with repetitions) is:
$9^{4}$
Each permutation of letters can be taken with each permutation of digits, therefore the total number of possible passwords is:
$25^{2}\times 9^{4}=4,100,625$

6 0

3 years ago

Read 2 more answers

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago