Change the mixed number 4 9/16 to an improper fraction

Write 0.941 with word name

jolli1 [7]

Nine tenths, four hundredths, and one thousandth.

OR:

Nine hundred and forty one thousandths.

7 0

4 years ago

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Carly placed 50 marbles in a bag. Fifteen of the marbles are pink, 8 marbles are black,

yuradex [85]

When you go into this problem, you want to figure out your marble ammount to 50 so in this case we will say C for color and 50 for the total ammount of marbles.

We know 15 are pink, 8 are black, 2 are green, 18 are clear, and 7 are striped

15P/50

8B/50

2G/50

18C/50

7S/50 for a total of 50 marbles

Now we use the chart to decide our awnsers

A. We know our propability of drawing a green and clear is 20/50 which if we simplify is a 2/5 ratio. If We put this in perspective 2/5 is rare and is unlikley to even.

B. We know a striped marble is 18/50 or 1.8/5 ratio which is mainly unlikely

C. We have 23/50 marbles that are black and pink, our propability is about 2.3/5 and gives us an even chance to get one of these

D. We know we have 33/50 marbles that are pink and clear and gives us a 3.3/5 chance of getting one of these and gives us an even to likely chance of getting one of these.

E. If we have a total of 17 marbles in these 3 colors, we have a 1.7/5 chance of getting one of these and is probably impossible to unlikey.

4 0

2 years ago

A printer toner company launches a new product. The number of pages that this new toner can print is normally distributed with a

Vitek1552 [10]

Answer:

Step-by-step explanation:

Since the number of pages that this new toner can print is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = the number of pages.

µ = mean

σ = standard deviation

From the information given,

µ = 2300 pages

σ = 150 pages

1)

the probability that this toner can print more than 2100 pages is expressed as

P(x > 2100) = 1 - P(x ≤ 2100)

For x = 2100,

z = (2100 - 2300)/150 = - 1.33

Looking at the normal distribution table, the probability corresponding to the z score is 0.092

P(x > 2100) = 1 - 0.092 = 0.908

2) P(x < 2200)

z = (x - µ)/σ/√n

n = 10

z = (2200 - 2300)/150/√10

z = - 100/47.43 = - 2.12

Looking at the normal distribution table, the probability corresponding to the z score is 0.017

P(x < 2200) = 0.017

3) for underperforming toners, the z score corresponding to the probability value of 3%(0.03) is

- 1.88

Therefore,

- 1.88 = (x - 2300)/150

150 × - 1.88 = x - 2300

- 288 = x - 2300

x = - 288 + 2300

x = 2018

The threshold should be

x < 2018 pages

4 0

3 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

4 years ago

Click on all the values for n that make this inequality true. There may be more than one answer.

vodka [1.7K]

Equation: 120 ≥ 10n + 20

Subtract 20 from both sides,

100 ≥ 10n

Divide both sides by 10,
10 ≥ n

It means, it is equal or Smaller than 10.

So, the answers would be: 7, 8, 9, 10

Hope this helps!

4 0

4 years ago

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