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3 years ago

8

What are the answers to the bottom two?

1 answer:

Ira Lisetskai [31]3 years ago

3 0

Step-by-step explanation:

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below ÷)

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Using appropriate properties find: 2/3 × 3/4 + 3/7 × 3/4

Viktor [21]

Answer:

$\frac{23}{28}$

Step-by-step explanation:

$\frac{2}{3} \times \frac{3}{4} + \frac{3}{7} \times \frac{3}{4}$

$= > \frac{3}{4} ( \frac{2}{3} + \frac{3}{7} )$

$= > \frac{3}{4} \times \frac{23}{21}$

Reducing 3 from numerator and denominator,

$= > \frac{23}{4 \times 7} = \frac{23}{28}$

3 0

2 years ago

There is a set of 12 cards numbered 1 to 12. A card is chosen at random. Event A is choosing a number less

kaheart [24]

The number of permutations of picking 4 pens from the box is 30.

There are six different unique colored pens in a box.

We have to select four pens from the different unique colored pens.

We have to find in how many different orders the four pens can be selected.

<h3>What is a permutation?</h3>

A permutation is the number of different arrangements of a set of items in a particular definite order.

The formula used for permutation of n items for r selection is:

$^nP_r = \frac{n!}{r!}$

Where n! = n(n-1)(n-2)(n-3)..........1 and r! = r(r-1)(r-2)(r-3)........1

We have,

Number of colored pens = 6

n = 6.

Number of pens to be selected = 4

r = 4

Applying the permutation formula.

We get,

= $^6P_4$

= 6! / 4!

=(6x5x4x3x2x1 ) / ( 4x3x2x1)

= 6x5

=30

Thus the number of permutations of picking 4 pens from a total of 6 unique colored pens in the box is 30.

Learn more about permutation here:

brainly.com/question/14767366

#SPJ1

6 0

9 months ago

out of 45 times at bat, Raul got 19 hits.Find Rauls batting average as a decimal rounded tothe nearst thousandth.

kati45 [8]

19/45=0.422222222........
Therefore rounded to the nearest thousandth his batting average is 0.422

3 0

2 years ago

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

3 years ago

Whoever answers the question first and right will get brainly

guapka [62]

Heyyyyyyydbdndnhbdbebe

4 0

2 years ago