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2 years ago

5

Lamar puts 3 cantaloupes that each weigh 2 pounds onto a scale. He adds 3 bunches of bananas that each have the same weight to t

he scale. If the total weight of the fruit is 18 pounds, what is the weight of one bunch of bananas? Write the equation and solve for the unknown.

2 answers:

butalik [34]2 years ago

7 0

The answer iss 4 => so yes

allochka39001 [22]2 years ago

6 0

4 is the weight of one bunch 6X4=18.

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The length of a rectangle is 4 cm less then twice it’s width the perimeter of the rectangle is 34 cm. What are the dimensions of

IRISSAK [1]

The lengths are 10 and the widthes are 7

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3 years ago

Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people

Leona [35]

Answer:

a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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.

In this problem we have that:

$p = 0.85, n = 6$ .

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

$P(X < 6) + P(X = 6) = 1$

$P(X < 6) = 1 - P(X = 6)$

In which:

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

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

So

$P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229$

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X \leq 4) = 1 - P(X > 4)$

In which

$P(X > 4) = P(X = 5) + P(X = 6)$

$P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993$

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764$

$P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236$

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

$P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960$

0.3771 + 0.3960 = 0.7764 = 77.64% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 8 donors

$P(X = 6) = C_{8,6}.(0.85)^{6}.(0.15)^{2} = 0.2376$

0.3771 + 0.3960 + 0.2376 = 1.01 = 101% probability of obtaining blood from at least six Rh+ donors over 0.95.

There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

5 0

3 years ago

Determine the time necessary for P dollars to double when it is invested at interest rate r compounded annually, monthly, daily,

kipiarov [429]

9514 1404 393

Answer:

annually: 9.01 years
monthly: 8.69 years
daily: 8.67 years
continuously: 8.66 years

Step-by-step explanation:

For interest compounded in discrete intervals, the formula is ...

A = P(1 +r/n)^(nt)

We want to find t for P=1 and A=2, so we have ...

2 = (1 +r/n)^(nt)

ln(2) = nt·ln(1+r/n)

t = ln(2)/(n·ln(1+r/n))

A table of values for r=0.08 is attached.

__

For continuous compounding, the formula is ...

A = Pe^(rt)

t = ln(A/P)/r = ln(2)/0.08 ≈ 8.66434 . . . . years

__

annually: 9.01 years
monthly: 8.69 years
daily: 8.67 years
continuously: 8.66 years

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2 years ago

How would you solve this: x^2-9x+18

Ostrovityanka [42]

Hope this helps you!!!

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago