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3 years ago

Can someone please help me on this problem?

Mathematics

1 answer:

uranmaximum [27]3 years ago

4 0

Answer:

$5820.10

Step-by-step explanation:

formula: $117\sqrt[3]{n^{2} }+1264$

n = 243

$117\sqrt[3]{243^{2} }+1264$ = 117(38.941)+ 1264

= 4556.097 + 1264

= 5820.10

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What is the probability that a five-card poker hand contains exactly one ace?.

spin [16.1K]

The probability of selecting exactly one ace is its likelihood

The probability that a five-card poker hand contains exactly one ace is 29.95%

<h3>How to determine the probability?</h3>

There are 4 aces in a standard deck of 52 cards.

The probability of selecting an ace would be:

p = 4/52

Also, there are 48 non-ace cards in the standard deck

So, the probability of selecting a non-ace after an ace has been selected is:

p = 48/51

The probability of selecting a non-ace up to the fifth selection are:

After two cards have been selected is: 47/50.
After three cards have been selected is: 46/49.
After four cards have been selected is: 45/48.

The required probability is then calculated as:

P(1 Ace) = n * (4/52) * (48/51) * (47/50) * (46/49) * (45/48)

Where n is the number of cards i.e. 5

So, we have:

P(1 Ace) = 5 * (4/52) * (48/51) * (47/50) * (46/49) * (45/48)

Evaluate

P(1 Ace) = 0.2995

Express as percentage

P(1 Ace) = 29.95%

Hence, the probability that a five-card poker hand contains exactly one ace is 29.95%

Read more about probability at:

brainly.com/question/25870256

8 0

3 years ago

Suppose u and v are functions of x that are differentiable at x=0 and that u(0)=4, u′(0)=7, v(0)=2 and v′(0)=1. Find the values

Leona [35]

Answer with Step-by-step explanation:

We are given that u and v are functions of x and both are differentiable at x=0

$u(0)=4,u'(0)=7,v(0)=2,v'(0)=1$

a.We have to find the values of $\frac{d(uv)}{dx}$

$\frac{d(u\cdot v)}{dx}=u'v+uv'$

Using this formula

Then , we get

$[\frac{d(uv)}{dx}]_{x=0}=u'(0)v(0)+u(0)v'(0)=7(2)+4(1)=14+4=18$

$[\frac{d(uv)}{dx}]_{x=0}=18$

b. $\frac{d(u/v)}{dx}=\frac{u'v-uv'}{v^2}$

$[\frac{d(u/v)}{dx}]_{x=0}=\frac{u'(0)v(0)-u(0)v'(0)}{v^2(0)}=\frac{7(2)-4(1)}{2^2}=\frac{14-4}{4}=\frac{10}{4}=\frac{5}{2}$

$[\frac{d(u/v)}{dx}]_{x=0}=\frac{5}{2}$

$[\frac{d(v/u)}{dx}]_{x=0}=\frac{v'(0)u(0)-v(0)u'(0)}{u^2(0)}=\frac{1(4)-7(2)}{4^2}$

$[\frac{d(v/u)}{dx}]_{x=0}=\frac{-10}{16}=\frac{-5}{8}$

d. $\frac{d(-6v-9u)}{dx}=-6v'-9u'$

$[\frac{d(-6v-9u)}{dx}]_{x=0}=-6v'(0)-9u'(0)=-6(1)-9(7)=-6-63=-69$

$[\frac{d(-6v-9u)}{dx}]_{x=0}=-69$

3 0

3 years ago

Solve for j. –9.3j + 12.34 = –14.64 − 5.2j − 6j

xeze [42]

So the value of j is -14.2

look at the attached picture

Hope it will help you

Good luck on your assignment

6 0

3 years ago

Read 2 more answers

A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure of

liq [111]

Answer:

The probability of failure of both the bulbs is 0.4323.

Step-by-step explanation:

For an exponential distribution the distribution is given by

$f(x,\lambda )=\int_{0}^{x }\lambda e^{-\lambda x}dx$

The value of λ is related to the mean μ as λ=1/μ,

Let us denote the 2 bulbs by X and Y thus the probability distribution of the 2 bulbs is as under

$P(X)=\int_{0}^{x }\lambda _{X}e^{-\lambda _{X}x}dx$

Similarly for the bulb Y the distribution function is given by

$P(Y)=\int_{0}^{y }\lambda _{Y}e^{-\lambda _{Y}y}dy$

Thus the probability for both the bulbs to fail within 1500 hours is

$P(E)=\int_{0}^{1500}\int_{0}^{1500}\frac{1}{1400}e^{\frac{-x}{1400}}\cdot \frac{1}{1400}e^{\frac{-y}{1400}}dxdy\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}\int_{0}^{1500}e^{\frac{-x}{1400}}\cdot e^{\frac{-y}{1400}}dxdy)\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}e^{\frac{-x}{1400}}dx)\cdot (\int_{0}^{1500}e^{\frac{-y}{1400}}dy)\\\\P(E)=\frac{1}{1400^{2}}\times 920.473\times 920.473\\\\\therefore P(E)=0.4323$

6 0

3 years ago

In graph, the area below f(x) is shaded and labeled A, the area below g(x) is shaded ad labeled B, and the area where f(x) and g

hjlf

Answer:

12 i did it

Step-by-step explanation:

6 0

3 years ago