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Yuliya22 [10]
3 years ago
6

Bananas are $0.59/pound.

Mathematics
2 answers:
gulaghasi [49]3 years ago
8 0
The answer for question B is $14.75
victus00 [196]3 years ago
6 0

Answer:

25 pounds will cost 14.75

Step-by-step explanation:

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What is the correct answer?
Aneli [31]

Answer:

b

Step-by-step explanation:

hhucudydydidydtvybi

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3 years ago
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Helena is in charge of selling tickets for the upcoming
iogann1982 [59]

Answer:

16

Step-by-step explanation:

If you add them all up you get 80. Then you divide it by 5 since there is 5 terms. 80/5 = 16

5 0
4 years ago
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Please hepl with math
Ivenika [448]

Answer:

\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}} = -\dfrac{5}{3}

Step-by-step explanation:

We are given the expression to be simplified:

\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}}

Let us take common a term with a power of 5 from the numerator and the denominator of the given expression.

We know that:

a^{p+q} = a^p \times a^q

Let us use it to solve the powers of 5 in the given expression.

\therefore we can write:

5^{n+2} = 5^{n+1}\times 5= 5^n\times 5^{2}

5^{n+1} = 5^n\times 5

The given expression becomes:

\dfrac{5^{n+1} \times 5-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n}\times 5}

Taking common 5^{n+1} from the numerator and

Taking common 5^{n} from the denominator

\Rightarrow \dfrac{5^{n+1} (5-6)} {5^{n}(13-2\times5)}\\\Rightarrow \dfrac{5^{n+1} (-1)} {5^{n}(13-10)}\\\Rightarrow -\dfrac{5^{n+1}} {5^{n}\times3}\\\Rightarrow -\dfrac{5^{n}\times 5} {5^{n}\times3}\\\Rightarrow -\dfrac{5}{3}

\therefore The answer is:

\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}} = -\dfrac{5}{3}

6 0
3 years ago
Two angles are complementary. One of the angles is twice the other.
sladkih [1.3K]
A + 2A = 180
3A = 180
A = 60

2A = 120A
7 0
4 years ago
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You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r
kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that p = 0.15

Assume you obtain a random sample of 9 individuals from this population:

This means that n = 9

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316

P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679

P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

5 0
3 years ago
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