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pantera1 [17]
2 years ago
13

A truck is carrying 10 3 bushels of apples, 10 2 bushels of grapes, and 10 1 bushels of oranges. Each bushel of apples weighs

32 pounds, each bushel of grapes weighs 25 pounds, and each bushel of oranges weighs 42 pounds. What is the total weight of the fruit in pounds?
Mathematics
1 answer:
almond37 [142]2 years ago
5 0

Answer:

10,088 pounds

Step-by-step explanation:

Given

103 bushels of apple

102 bushels of grape

101 bushels of oranges

Also weight of each bushel:

Apple = 32 pounds

Grape = 25 pounds

Oranges = 42 pounds

Total weight will be sum of product of each:

Total Weight: 103(32) + 102(25) + 101(42) = 10,088 pounds

Guest
1 year ago
Thanks
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Thomas graphed the line that represents the equation y=34x.
zloy xaker [14]

Answer:

The ordered pairs represent points on the line are

(4, 3) ⇒ C

(2, \frac{3}{2} ) ⇒ D

(-8, -6) ⇒ E

Step-by-step explanation:

To find the ordered pairs represent points on the line, substitute x by the x-coordinate of each point, if the value of y equals the y-coordinate of the point, then the point is on the line.

∵ The equation is y = \frac{3}{4} x

∵ The ordered pair is (8, \frac{1}{6} )

→ Substitute x by 8

∴ y = \frac{3}{4} (8)

∴ y = 6

∵ The value of y does not equal the y-coordinate of the ordered pair

∴ The ordered pair (8, \frac{1}{6} ) does not represent a point on the line

∵ The ordered pair is (\frac{-2}{3}, \frac{1}{2} )

→ Substitute x by \frac{-2}{3}

∴ y = \frac{3}{4} (\frac{-2}{3})

∴ y = \frac{-1}{2}

∵ The value of y does not equal the y-coordinate of the ordered pair

∴ The ordered pair  (\frac{-2}{3}, \frac{1}{2} ) does not represent a point on the line

∵ The ordered pair is (4, 3 )

→ Substitute x by 4

∴ y = \frac{3}{4} (4)

∴ y = 3

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (4, 3) represents a point on the line

∵ The ordered pair is (2, \frac{3}{2} )

→ Substitute x by 2

∴ y = \frac{3}{4} (2)

∴ y = \frac{3}{2}

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (2, \frac{3}{2} ) represents a point on the line

∵ The ordered pair is (-8, -6 )

→ Substitute x by -8

∴ y = \frac{3}{4} (-8)

∴ y = -6

∵ The value of y equal the y-coordinate of the ordered pair

∴ The ordered pair (-8, -6) represents a point on the line

5 0
1 year ago
The heights of 1000 students are approximately normally distributed with a mean of 174.5 centimeters and a standard deviation of
yuradex [85]
A. The mean and standard deviation.

The mean of a sampling distribution is approximately equal to the mean of the population. Given that the mean of the population is equal to 174.5, the mean of the sampling distribution is also this value.

The standard deviation of a sample distribution is equal to,

                u(m) = u/sqrt n

Substituting the known values,

               u(m) = 6.9 / sqrt 25 = 1.38

b. Get the z-score of both items,
         
      z-score = (data point - mean) / standard deviation

 z-score of 172.5
     z-score = (172.5 - 174.5) / 1.38 = -1.49
This translates to 0.068.

z-score of 175.8
   z-score = (175.8 - 174.5) / 1.38 = 0.94
This translates to 0.83. 

The difference between the two z-scores is 0.762. 

 The number of samples with this height is 0.762(200) which is equal to approximately 152.

c. z-score of 172 centimeters
   
    z-score = (172 - 174.5) / 1.38
  
    z-score = -1.81
This translates to 0.03.

The number of people with this height from the sample is (0.03)(200) = 6
8 0
2 years ago
In high-school 135 freshmen were interviewed.
timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

3 0
1 year ago
Read 2 more answers
A student theater charges $9.50 per ticket. The theater has already sold 70 tickets. Write and solve an inequality that represen
Genrish500 [490]
9.50 times 70 is greater or less than 1000
8 0
2 years ago
Two cross sections of a right hexagonal pyramid are obtained by cutting the pyramid with planes parallel to the hexagonal base.
Tanya [424]

Answer:

The larger cross section is 24 meters away from the apex.

Step-by-step explanation:

The cross section of a right hexagonal pyramid is a hexagon; therefore, let us first get some things clear about a hexagon.

The length of the side of the hexagon is equal to the radius of the circle that inscribes it.

The area is

A=\frac{3\sqrt{3} }{2} r^2

Where r is the radius of the inscribing circle (or the length of side of the hexagon).

Now we are given the areas of the two cross sections of the right hexagonal pyramid:A_1=216\:ft^2\: \:\:\:A_2=486\:ft^2

From these areas we find the radius of the hexagons:

r_1=\sqrt{\frac{2A_1}{3\sqrt{3} } } =\sqrt{\frac{2*216}{3\sqrt{3} } }=\boxed{9.12ft}

r_2=\sqrt{\frac{2A_2}{3\sqrt{3} } } =\sqrt{\frac{2*486}{3\sqrt{3} } }=\boxed{13.68ft}

Now when we look at the right hexagonal pyramid from the sides ( as shown in the figure attached ), we see that r_1 r_2 form similar triangles with length H

Therefore we have:

\frac{H-8}{r_1} =\frac{H}{r_2}

We put in the numerical values of r_1, r_2 and solve for H:

\boxed{H=\frac{8r_2}{r_2-r_1} =\frac{8*13.677}{13.68-9.12} =24\:feet.}

8 0
2 years ago
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