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kirill115 [55]
3 years ago
8

Noah and Lin are making paper cones to hold popcorn to hand out at parent math night. They want the cones to hold 9 cubic inches

of popcorn. What are two different possible values for height h and radius r for the cones?
height (h)=

radius (r)=
Mathematics
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

So, The possible values of h and r are

1. r = 3 , h = 3

2. r = 1, h = 27

3. r = 2, h = 6.75

Step-by-step explanation:

P.S - The exact ques is -

As we know that Volume of cone , V = \frac{1}{3}\pi r^{2}h

As given,

V = 9\pi

⇒\frac{1}{3}\pi r^{2}h = 9\pi \\  \frac{1}{3} r^{2}h = 9 \\r^{2}h = 27

So,

The possible values of h and r are

1. r = 3 , h = 3

2. r = 1, h = 27

3. r = 2, h = 6.75

In Case I-

When r = 3, h = 3

(3)² ×3 = 9×3 = 27

Satisfied

Case II-

r = 1, h = 27

(1)²×27 = 1×27 = 27

Satisfied

Case III-

r = 2, h = 6.75

(2)²×6.75 = 4×6.75 = 27

Satisfied

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(5 2/7 −1 1/2 )∙(1− 14/53 )
yulyashka [42]

I'll teach and give you the answer for you problem (5 2/7 −1 1/2 )∙(1− 14/53 )

Convert mixed numbers to improper fractions 5 2/7 = 37/7:

(37/7-1 1/2) * (1- 14/53

Convert mixed numbers to improper fractions 1 1/2 = 3/2:

(37/7-3/2) * (1- 14/53

Join 37/7-3/2 = 53/14

53/14(-14/53+1)

Join 1-14/53 = 39/53

53/14 * 39/53

Cross-Cancel 53:

39/14

Your Answer As A Improper Fraction---> 39/14

Your Answer As A Mixed Number---> 2  11/14

plz mark me as brainliest :)

7 0
3 years ago
Classify the following statement as an example of classical​ probability, empirical​ probability, or subjective probability. Exp
AlekseyPX

Answer:

It is a case of classical probability.

Step-by-step explanation:

Since we are selecting a number between 1 and 100 randomly which is divisible by 14 the only favorable cases are 14,28,42,56,70,84,98 which are 7 cases out of 100 total numbers thus the required probability becomes

P(E)=\frac{7}{100}=0.07

8 0
3 years ago
There are 50 deer in a particular
olchik [2.2K]

answered

There are 50 deer in a particular forest. The population is increasing at a rate of 15% per year. Which exponential growth function represents

the number of deer y in that forest after x months? Round to the nearest thousandth.

1

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dribeiro

Ace

697 answers

561.4K people helped

Answer:

The expression that represents the number of deer in the forest is

y(x) = 50*(1.013)^x

Step-by-step explanation:

Assuming that the number of deer is "y" and the number of months is "x", then after the first month the number of deer is:

y(1) = 50*(1+ 0.15/12) = 50*(1.0125) = 50.625

y(2) = y(1)*(1.0125) = y(0)*(1.0125)² =51.258

y(3) = y(2)*(1.0125) = y(0)*(1.0125)³ = 51.898

This keeps going as the time goes on, so we can model this growth with the equation:

y(x) = 50*(1 - 0.15/12)^(x)

y(x) = 50*(1.013)^x

8 0
2 years ago
Which of the following sets could not be the lengths of the sides of a triangle?
frez [133]
-(11,14,25) is the correct answer. the two shorter sides must add to a longer distance than the longest side.

8 0
2 years ago
A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular custome
Ludmilka [50]

Answer:

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), <em>p</em> = 0.52

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.52.

(1)

Compute the probability of overbooking  as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

                                        =P(X>3)\\\\={5\choose 4}(0.52)^{4}(1-0.52)^{5-4}+{5\choose 5}(0.52)^{5}(1-0.52)^{5-5}\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

=P(X

Thus, the probability that the flight has empty seats is 0.4625.

4 0
3 years ago
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