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BabaBlast [244]

2 years ago

5

If a cooler contains six cola, two grape, two orange, and four lemon-lime sodas, what is the probability of selecting a cola, gi

ving it to your friend, then selecting another cola for yourself?

1 answer:

frutty [35]2 years ago

6 0

Answer:

ok so first of all there is 14 sodas in the cooler(6+2+2+4)

and 6 of these 14 are cola 6/14 or 3/7

so 3/7 then to give your self one we multiple it by 3/7

3/7*3/7=0.18367346938

we multiple by 10

18.367346938

so the probility is 18.37%

Hope This Helps!!!

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Need Answer to these questions. pls help

vredina [299]

Answer:

7.2 million dollars

Step-by-step explanation:

The domain of the function the complete set of possible values of the independent variable. In this case the domain includes all real numbers except 189.

From;

T(p) = 0.50 (p - 189)

When p = 203.4 million dollars

T(p) = 0.50 (203.4 - 189)

T(p) = 7.2 million dollars

8 0

2 years ago

Amy bought a diamond ring for $6,000. If the value

Temka [501]

Answer: $13,308

5.8% of 6000 is 348. 348 times 21 is $7,308. add 6,000 to 7,308 and you get $13,308

Step-by-step explanation: amy likes cheap rings LOL

3 0

2 years ago

Read 2 more answers

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

2 years ago

Which word should replace the question mark (?) in the

artcher [175]

Answer:

speed I'm pretty sure sorry if its wrong

7 0

3 years ago

Read 2 more answers

ANSERR GIVE ME ANSWERS PLS

Zarrin [17]

Answer:

Concept: Mathematics

When you multiply a number by something to the 10 power of something then you move whatever place of the exponential to the right if its positive and left if it's negative.

1.365 × 10^1
You see its positive 1 hence we move one space to the right
Our answer is now 13.65

You keep doing the same thing

1.365 × 10^2
136.5

Lastly

1.365 ×10^3
1365

Rate positively and give 5 stars

4 0

3 years ago