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slavikrds [6]
3 years ago
10

To give their animals essential minerals and nutrients, farmers and ranchers often have a block of salt—called “salt lick”—avail

able for their animals to lick. The "salt lick" is a cube with an edge length of 5/12 ft.
b. The box that contains the salt lick is 1 1/4 feet by 1 2/3 feet by 5/6 feet. How many cubes of salt lick fit in the box.
Mathematics
1 answer:
Anna007 [38]3 years ago
5 0

Answer:

32 i think dont blame me if im way of

Step-by-step explanation:

You might be interested in
A rectangular patio has a length of 12 1/2 feet and an area of 103 1/8 square feet. What is the width of the patio?
Rina8888 [55]
Length of the rectangular patio = 12 1/2 feet
                                                  = 25/2 feet
Area of the rectangular patio = 103 1/8 square feet
                                               = 825/8 square feet
Let us assume the width of the rectangular patio = x feet
 Then
Area of the rectangular patio = Length * Width
825/8 = (25/2) * x
25x/2 = 825/8
25x = (825 * 2)/8 feet
25x = 825/4 feet
x = 825/(4 * 25) feet
   = 33/4 feet
   = 8 1/4 feet
So the width of the rectangular patio is 8 1/4 feet. I hope the procedure is clear enough for you to understand.
6 0
3 years ago
Which formula should be used to find the circumference of a circle
aleksandrvk [35]


If a circle has radius
R, its circumference equals to
2πR where π is an irrational number that, approximately, equals to
3.1415926

7 0
3 years ago
The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development
Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

X \sim Bin(n =20, p=0.8)

b) X \sim Bin(n =20, p=0.8)

c) P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

d) P(X \geq 15) = P(X=15)+ .....+P(X=20)

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012

P(X\geq 15)=0.804208

e) E(X) = np = 20*0.8 = 16

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

X \sim Bin(n =20, p=0.8)

Part b

X \sim Bin(n =20, p=0.8)

Part c

For this case we just need to replace into the mass function and we got:

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

Part d

For this case we want this probability: P(X\geq 15)

And we can solve this using the complement rule:

P(X \geq 15) = P(X=15)+ .....+P(X=20)

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012

P(X\geq 15)=0.804208

Part e

The expected value is given by:

E(X) = np = 20*0.8 = 16

5 0
3 years ago
{y=-x+1<br> {y=4x-14<br> please help me <br> my paper is due today
Temka [501]

- x + 1 = 4x - 14 \\  - x - 4x =  - 14 - 1 \\  - 5x =  - 15 \\ x = 3

Substitute x = 3 in any given equations. I will do in y = - x + 1

y =  - x + 1 \\ y =  - 3 + 1 \\ y =  - 2

Therefore the answer is (3,-2)

7 0
3 years ago
The final scores from the last five games of two basketball teams are given. Jumping Jackrabbits: 70, 65, 72, 80, 73 Leaping Liz
scoundrel [369]

Using the Mean Absolute Deviation(MAD) concept, it is found that:

a) The Jumping Jackrabbits have a MAD of 3.6 and the Leaping Lizards of 6.8.

b) The MAD is lower for the Jumping Jackrabbits then for the Leaping Lizards, hence their scores are more consistent.

<h3>What is the mean absolute deviation of a data-set?</h3>

  • The mean of a data-set is given by the sum of all observations divided by the number of observations.
  • The mean absolute deviation(MAD) of a data-set is the sum of the absolute value of the difference between each observation and the mean, divided by the number of observations.
  • The mean absolute deviation represents the average by which the values differ from the mean.

Item a:

For the Jumping Jackrabbits, the mean and MAD are given by:

M = (70 + 65 + 72 + 80 + 73)/5 = 72

MAD = (|70 - 72| + |65 - 72| + |72 - 72| + |80 - 72| + |73 - 72|)/5 = 3.6.

For the Leaping Lizards, the mean and the MAD are given by:

M = (61 + 47 + 70 + 63 + 54)/5 = 59

MAD = (|61 - 59| + |47 - 59| + |70 - 59| + |63 - 59| + |54 - 59|)/5 = 6.8.

The Jumping Jackrabbits have a MAD of 3.6 and the Leaping Lizards of 6.8.

Item b:

The MAD is lower for the Jumping Jackrabbits then for the Leaping Lizards, hence their scores are more consistent.

More can be learned about the Mean Absolute Deviation(MAD) concept at  brainly.com/question/3250070

#SPJ1

4 0
2 years ago
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