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3 years ago

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A space for storing boxes is 36 inches high. Each box is 6 inches high. A space of 9 inches must be left at the top. what is the

maximum number of boxes that can be stacked in this space?

1 answer:

Dvinal [7]3 years ago

7 0

Answer:

Step-by-step explanation: 36-9 equals 27. You can only fit 4 evenly so thats the answer

You might be interested in

The circumference of ruby's old swimming pool is 47.1 feet. Her new swimming pool has a diameter of 21 feet. How much larger is

Alla [95]

The new pool is 18.84 feet larger in circumference than the old one.

<u>Explanation:</u>

Given:

Circumference of the pool = 47.1 feet

Diameter of the new pool, d = 21 feet

radius, r = 21/2 feet

Circumference of the new pool = ?

We know:

Circumference = 2πr

where,

r is the radius

On substituting the value:

C = $2 X 3.14 X \frac{21}{2}$

C = 65.94 feet

The difference in circumference of the two pools = 65.94 - 47.1 feet

= 18.84 feet

Therefore, the new pool is 18.84 feet larger in circumference than the old one.

6 0

3 years ago

Find a unit vector that has the same direction as the vector a and the opposite direction of the vector

nasty-shy [4]

The unit vector is given by the following formula:
a '= (a) / (lal)
Where,
a: vector a
lal: Vector module a
We are looking for the module:
lal = root ((- 15) ^ 2 + (8) ^ 2)
lal = 17
Same direction:
a = -15i + 8j
The unit vector is:
a '= (1/17) * (- 15i + 8j)
Opposite direction:
a = 15i - 8j
The unit vector is:
a '= (1/17) * (15i - 8j)
Answer:
a unit vector that has the same direction as the vector a is:
a '= (1/17) * (- 15i + 8j)
a unit vector that has the opposite direction of the vector a is:
a '= (1/17) * (15i - 8j)

5 0

3 years ago

What is the fourth term of an arithmetic sequence whose first term is 23 and whose seventh term is 5?

mixas84 [53]

We have that the fourth term of an arithmetic sequence is

a_4=14

Option C

From the question we are told

What is the fourth term of an arithmetic sequence whose first term is 23 and whose seventh term is 5?

A) 78

B) 32

C) 14

(Explain your work)

Generally the equation for the arithmetic sequence is mathematically given as

$a_4=a_1+(n-1)d$

Therefore

For seventh term

$5=23+(6)d\\\\d=5-23/(6)\\\\d=-3$

Therefore

For Fourth term

$a_4=23+(3*(-3))$

a_4=14

For more information on this visit

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4 0

3 years ago

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n

Ivanshal [37]

Answer:

<u><em>a) The probability that exactly 4 flights are on time is equal to 0.0313</em></u>

<u><em></em></u>

<u><em>b) The probability that at most 3 flights are on time is equal to 0.0293</em></u>

<u><em></em></u>

<u><em>c) The probability that at least 8 flights are on time is equal to 0.00586</em></u>

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

<em>United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures. </em>

<em>a) The probability that exactly 4 flights are on time is = </em>

<em>b) The probability that at most 3 flights are on time is = </em>

<em>c)The probability that at least 8 flights are on time is =</em>

<h2>Solution to the problem</h2>

<u><em>a) Probability that exactly 4 flights are on time</em></u>

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

$P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}$

Where $_nC_x$ is the number of different combinations of x success in n trials.

$_nC_x=\frac{x!}{n!(n-x)!}$

Hence,

$P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}$

$_9C_4=\frac{4!}{9!(9-4)!}=126$

$P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125$

<em><u>b) Probability that at most 3 flights are on time</u></em>

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=(0.5)^9=0.00195313$ . . . (the probability that all are not on time)

$P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625$

$P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125$

$P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625$

$P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\ P(X\leq 3) \approx 0.0293$

<em><u>c) Probability that at least 8 flights are on time </u></em>

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

$P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586$

6 0

3 years ago

A town has a population of 19000 and grows at 4.5% every year. To the nearest tenth of a year, how long will it be until the pop

steposvetlana [31]

Answer:

If the population grows by 4% each year then

the population in any given year is 104% of

the previous year or 1.04 times as much

P(t) = P0(1.04)t

P(t) is population at time t years

P0 = initial population = 11,000

t = number of years = 15

P(15) = 11,000(1.04)15

P(15) = 19,810 people

Step-by-step explanation:

3 0

3 years ago