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2 years ago

7

matthew is planning dinners for the next 3 nights. there’s are 11 meals to choose from. If no meal is repeated, how many differe

nt meal arrangements are possible?

1 answer:

S_A_V [24]2 years ago

5 0

Answer:

may be I am not sure 11

Step-by-step explanation:

i think ok just a guess

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What is 63.4 divided by 20

Zanzabum

317 is the answer you are looking for.

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2 years ago

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According to the manufacturer, about 26% of sour candy in a package of Sandy's Sours are grape. What is the probability that the

Leto [7]

Answer:

Step-by-step explanation:

We assume that there are 100 sour candies, Thus-

26 % of candy are grape implies that 26% of 100 candies are grape that is equal to 26

Now remaing candies that are not grape are 100-26 = 74

Based on the rule of multiplication:

P(A ∩ B) = P(A)/ P(B|A)

 In the beginning, there are 26grape candies, probability of choosing first grape candy = 26C1 = 26

 After the first selection, we replace the selected grape candy so there are still 100 candies in the bag P(B|A) = 100C3 = 100 x 99 x 98 x 97!/3! X 97!

= 50 x 33 x 98

So probability =1/ 50 x 33 x 98 x 26

= 1/4204200

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2 years ago

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6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the

MaRussiya [10]

Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

$a_1$ the horizontal distance between the first observer and the ballonist

$a_2$ the horizontal distance between the second observer and the ballonist

$\alpha _1$ and $\alpha _2$ the angles of elevation meassured by each observer

S the area of the triangle formed with the observers and the ballonist

So, the area of a triangle is the length of its base times its height.

$S=a*h$ (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

$S=S_1+S_2$ (equation 2)

$S_1=a_1*h$

But we can write $S_1$ in terms of $\alpha _1$ , like this:

$\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}$

And for $S_2$ will be the same:

$S_2=\frac{h^{2} }{\tan(\alpha _2)}$

Replacing in the equation 2:

$S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})$

And replacing in the equation 1:

$h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}$

So, we can replace all the known data in the last equation:

$h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft$

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

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2 years ago

22.Lenin is preparing dinner plates. He has 12 pieces of chicken and 16 rolls. If he wants to make all the plates identicalwitho

Nitella [24]

Answer: The greatest number of plates Lenin can prepare = 4

and there will be 3 chickens and 4 rolls in each plate.

Step-by-step explanation:

Given: Lenin is preparing dinner plates. He has 12 pieces of chicken and 16 rolls.

To make all the plates identical without any food left over, the greatest number of plates Lenin can prepare = G.C.D.(12,16)=4

The number of pieces of chicken in each plate = $\dfrac{12}{4}=3$

The number of pieces of rolls in each plate = $\dfrac{16}{4}=4$

So, the greatest number of plates Lenin can prepare = 4

and there will be 3 chickens and 4 rolls in each plate.

7 0

2 years ago

See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly se

REY [17]

Answer:

a) There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b) 0.0064% probability that all of the selected adults believe in reincarnation.

c) There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d) Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

Step-by-step explanation:

For each of the adults selected, there are only two possible outcomes. Either they believe in reincarnation, or they do not. So 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.

In this problem we have that:

$n = 6, p = 0.2$

a. What is the probability that exactly five of the selected adults believe in reincarnation?

This is P(X = 5).

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

$P(X = 5) = C_{6,5}.(0.2)^{5}.(0.8)^{1} = 0.0015$

There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.

b. What is the probability that all of the selected adults believe in reincarnation?

This is P(X = 6).

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

$P(X = 6) = C_{6,6}.(0.2)^{6}.(0.8)^{0} = 0.000064$

There is a 0.0064% probability that all of the selected adults believe in reincarnation.

c. What is the probability that at least five of the selected adults believe in reincarnation?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564$

There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.

d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation?

5 is significantly high if $P(X \geq 5) < 0.05$

We have that

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564 < 0.05$

Since $P(X \geq 5) < 0.05$ , 5 is a significantly high number of adults who believe in reincarnation in this sample.

5 0

2 years ago