MATHEMATICS COLLEGE

(Giving brainliest!!)Terrance packs two boxes inside a crate. The total weight of the crate and two boxes is 98.45 pounds. One box weighs 32.8 pounds, and the other box weighs 39.34 pounds. How much does the crate weigh?

A) 72.14 pounds
B) 59.11 pounds
D) 26.31 pounds

Answers

Answer 1
Answer:

If the total weight of the crate and two boxes is 98.45 pounds and one box weighs 32.8 pounds, and the other box weighs 39.34 pound, the weight of the crate is D) 26.31 pounds.

How the weight of the crate is determined:

The weight of the crate can be determined by addition and subtraction.

Addition and subtraction involves the two of the four basic mathematical operations, including division and multiplication.

The total weight of the crate and two boxes = 98.45 pounds

The weight of one box = 32.8 pounds

The weight of the second box = 39.34 pounds

The total weight of the boxes = 72.14 pounds (32.8 + 39.34)

The weight of the crate = 26.31 pounds (98.45 - 72.14)

Learn more about addition and subtraction at brainly.com/question/4721701.

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Answer 2
Answer:

Answer:

the crate weighs 26.31

Step-by-step explanation:

add 32.8 and 39.34 together and then subtract 98.45 from the sum. ^^

Step-by-step explanation:

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Three out of every ten dentists recommend a certain brand of fluoride toothpaste. Which assignment of random digits would be used to simulate the random sampling of dentists who prefer this fluoride toothpaste?A. none of theseB. Let “1,2,3” represent preferring the toothpaste and “0,4,5,6,7,8,9” represent not recommending the toothpaste.C. Let “1,2,3” represent preferring the toothpaste and “4,5,6,7,8,9” represent not recommending the toothpaste.D. Let “0,1,2” represent preferring the toothpaste and “3,4,5,6,7,8” represent not recommending the toothpaste.Reset Selection

What is 4901 and 75 estimated to

Answers

on what scale? to the nearest what? Elaborations, PLEASE

Anyways it's just:

4901 (4900*) + 75 (80*) = 4980

*i'm guessing

BUT if they want you to round after addition:

the answer would be 5000

Another rock in Death Valley at the Racetrack traveled 55.3 meters during the observation period of nine years. a . What was the rock's average speed in feet per year? Round your answer to the nearest whole number. feet per year b . If a strong wind gives a stone an initial velocity of v 0 = 11 m / s , and the friction coefficient of stone on ice is μ = 0.14 , and g = 9.81 m / s 2 . Use the equation v 2 0 = 2 μ g S to find the stopping distance S for the stone. Round your answer to the nearest tenth.

Answers

Answer:

a) 20 feet per year

b) 44.1 m

Step-by-step explanation:

Given:

Distance traveled during the observation period = 55.3 meters

Observation period = 9 years

initial velocity of v₀ = 11 m/s

friction coefficient of stone on ice is μ = 0.14

g = 9.81 m/s²

also,

v₀² = 2μgS

Now,

1 m = 3.28084 ft

thus,

Total distance in feet = 55.3 × 3.28084 = 181.430452 ft

Average speed = \frac{\textup{Distance}}{\textup{Time}}

or

Average speed = \frac{\textup{181.430452}}{\textup{9}}

or

Average speed = 20.159 feet/year ≈ 20 feet per year

b) v₀² = 2μgS

substituting the values in the above equation, we get

11² = 2 × 0.14 × 9.81 × s

or

121 = 2.7468 × s

or

s = 44.051 ≈ 44.1 m

Researchers suspect that the average number of units earned per semester by college students is rising. A researcher at Calendula College wishes to estimate the number of units earned by students during the spring semester at Calendula. To do so, he randomly selects 100 student transcripts and records the number of units each student earned in the spring term. He found that the average number of semester units completed was 12.96 units per student. Identify the population of interest to the researcher.

Answers

Answer:

All Calendula College students enrolled in the spring.

Step-by-step explanation:

A researcher at Calendula College wishes to estimate the number of units earned by students during the spring semester at Calendula.

To do so, he randomly selects 100 student transcripts from among all Calendula College students enrolled in the spring and records the number of units each student earned in the spring term.

A piece of rope there is 28 feet long is cut into two pieces. One is use to form a circle and others used to form a square. Write a function G representing the area of the square as a function of the radius of the circle R

Answers

The function g representing the area of the square as a function of the radius of the circle r is given as:

g(r) = 49 - 22r + (121r^2)/(49)

Solution:

Given that,

length of rope = 28 feet

Let "c" be the circumference of circle

Let "p" be the perimeter of square

Therefore,

length of rope = circumference of circle + perimeter of square

c + p = 28 ------- eqn 1

The circumference of circle is given as:

c = 2 \pi r

Where, "r" is the radius of circle

Substitute the above circumference in eqn 1

2 \pi r + p = 28

p = 28 - 2 \pi r ----------- eqn 2

If "a" is the length of each side of square, then the perimeter of sqaure is given as:

p = 4a

Substitute p = 4a in eqn 2

4a = 28 - 2 \pi r\n\na = (28 - 2 \pi r)/(4)\n\na = 7 - ( \pi r)/(2)

The area of square is given as:

area = (side)^2\n\narea = a^2

Substitute the value of "a" in above area expression

area = (7 - ( \pi r)/(2))^2  ------ eqn 3

We know that,

(a - b)^2 = a^2 - 2ab + b^2

Therefore eqn 3 becomes,

area = 7^2 -2(7)((\pi r)/(2)) + (( \pi r)/(2))^2\n\narea = 49 - 7 \pi r + ( (\pi)^2 r^2 )/(4)

\text{ substitute } \pi = (22)/(7)

area = 49 - 7 * (22)/(7) * r + ((22)/(7))^2 * (r^2 )/(4)\n\narea = 49 - 22r + (121)/(49) * r^2\n\narea = 49 - 22r + (121r^2)/(49)

Let g(r) represent the area of the square as a function of the radius of the circle r, then we get

g(r) = 49 - 22r + (121r^2)/(49)

Thus the function is found

I need help solving this equatioion.
(2xy)(3y)

Answers

Step-by-step explanation:

So first we will multiply

(2xy)(3y)

=

6x {y}^(2)

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experience that 1%, 3%, and 2% of the products made by each machine, respectively, are defective. A finished product is randomly selected and found to be non-defective, what is the probability that it was made by machine B1?

Answers

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = (P(B|A)P(A))/(P(B)) = (P(B|A)P(A))/(P(B|A)P(A) + P(B|a)P(a))

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =(P(N|B1)P(B1))/(P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)) = ((0.297)(0.3))/((0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)) = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.
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