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3 years ago

Chris sells roses for $25.99 a bunch. At the end of the day he has collected $285.89. How many bunches of roses did he sell

Mathematics

2 answers:

kondaur [170]3 years ago

8 0

He sells 11 bunches of roses.

Kryger [21]3 years ago

3 0

11 bunches
In order to find how many bunches he sold, do 285.89/25.99
hope this helps!

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3 years ago

a) Γ(n) =∫[infinity]0tn−1e−tdta)Show that Γ(n+ 1) =nΓ(n) using integration by parts.b) Show that Γ(n+ 1) =n! , wherenis a positi

Nataliya [291]

$\Gamma(n)=\displaystyle\int_0^\infty t^{n-1}e^{-t}\,\mathrm dt$

a. Integrate by parts by taking

$u=t^{n-1}\implies\mathrm du=(n-1)t^{n-2}\,\mathrm dt$

$\mathrm dv=e^{-t}\,\mathrm dt\implies v=-e^{-t}$

Then

$\displaystyle\Gamma(n)=-t^{n-1}e^{-t}\bigg|_0^\infty+(n-1)\int_0^\infty t^{n-2}e^{-t}\,\mathrm dt$

We have

$\displaystyle\lim_{t\to\infty}\frac{t^{n-1}}{e^t}=0$

and so

$\Gamma(n)=(n-1)\displaystyle\int_0^\infty t^{n-2}e^{-t}\,\mathrm dt=(n-1)\Gamma(n-1)$

or, replacing $n\to n+1$ , $\Gamma(n+1)=n\Gamma(n)$ .

b. From the above recursive relation, we find

$\Gamma(n+1)=n\Gamma(n)=n(n-1)\Gamma(n-1)=n(n-1)(n-2)\Gamma(n-2)=\cdots=n(n-1)(n-2)\cdots2\cdot1\Gamma(1)$

Now,

$\Gamma(1)=\displaystyle\int_0^\infty e^{-t}\,\mathrm dt=1$

and so we're left with $\Gamma(n+1)=n!$ .

c. Using the previous result, we find $\Gamma(5)=4!=24$ .

d. If the question is asking to find $\Gamma(12)$ , then you can just use the same approach as in (c).

But if you're supposed to find $\Gamma\left(\frac12\right)$ , we have

$\displaystyle\Gamma\left(\frac12\right)=\int_0^\infty t^{-1/2}e^{-t}\,\mathrm dt$

Substitute

$u=t^{1/2}\implies u^2=t\implies 2u\,\mathrm du=\mathrm dt$

Then

$\displaystyle\Gamma\left(\frac12\right)=\int_0^\infty\frac1ue^{-u^2}(2u\,\mathrm du)=2\int_0^\infty e^{-u^2}\,\mathrm du=\frac{2\sqrt\pi}2=\sqrt\pi$

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3 years ago

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Mrac [35]

False because y is not shown in the Equation

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2 years ago

What is an

miv72 [106K]

Answer:

y=1/4x + 1

Step-by-step explanation:

first equation,,y=1/4x - 1

G1 = G2

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4y = X + 8 -4

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2 years ago

f a random sample of 300 adults is taken from the state of Colorado, where the rate of obesity is 19.8%, can we use the Normal a

Troyanec [42]

Answer:

Probability of at least 50 obese individuals in our sample is 0.92364 .

Step-by-step explanation:

We are given that a random sample of 300 adults is taken from the state of Colorado, where the rate of obesity is 19.8% .

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For approximating binomial distribution into normal distribution, firstly we have to calculate $\mu$ and $\sigma^{2}$ .

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Variance of Normal distribution, $\sigma^{2}$ = n * p * (1-p) = 300 *0.198 *0.802 = 47.64

So, now X ~ N( $\mu = 59.4 , \sigma^{2} = 47.64$ )

The standard normal z score distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

So, probability of at least 50 obese individuals in our sample = P(X >= 50)

P(X >= 50) = P(X > 49.5) {using continuity correction}

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Therefore, required probability is 0.92364 .

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3 years ago