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3 years ago

One of the large pumpkins ever grown weighed about 3/4 ton. How many pounds did the pumpkins weigh ?

Mathematics

2 answers:

Oksana_A [137]3 years ago

6 0

One ton is 2000 lbs (pounds). 3/4 of 2000 is 1500, therefore the pumpkin weighed 1500 pounds.

Hunter-Best [27]3 years ago

4 0

The correct answer is 1500 pounds.
One ton is equal to 2000 pounds. You just divide 2000 by four and multiply that answer by three.

You might be interested in

The density of a certain material is such that it weighs 9 kilograms for every 7 cubic feet of volume. Express this density in p

Vadim26 [7]

Answer:

1.641× 10^-3 pounds/ cubic inch

Step-by-step explanation:

Density= mass / volume

Mass of the material= 9 kilograms

Volume= 7 cubic feet

Density= 9/7= 1.286 kg/ ft^3

Density= 1.286 kg/ ft^3

To Convert to pounds per inch

1 kg/ ft^3 = 0.00128 pound / cubic inch

1.286 kg/ ft^3 = X pounds/ cubic inch

X=( 1.286 × 0.00128)

= 0.0016407087 pounds/ cubic inch

= 1.641× 10^-3 pounds/ cubic inch

Hence,this density in pounds per inch is 1.641× 10^-3 pounds/ cubic inch

5 0

2 years ago

Explain how to solve this problem:

galina1969 [7]

"every dollar" means P = 1

compounded monthly interest means 12

1(1+(.129/12)^(12(1)) =

in calculator do pemdas

1. .129/12 = 0.01075

2. 0.01075 + 1 = 1.01075

3. (1.01075)^12 = 1.136907156 =

answer = 1.14

Mayfield schools

lcps

7 0

2 years ago

an outlier may be defined as a data point that is more than 1.5 times the interquartile range below the lower quartile or is mor

Ira Lisetskai [31]

Supposing a normal distribution, we find that:

The diameter of the smallest tree that is an outlier is of 16.36 inches.

--------------------------------

We suppose that tree diameters are normally distributed with mean 8.8 inches and standard deviation 2.8 inches.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

Mean of 8.8 inches, thus $\mu = 8.8$ .
Standard deviation of 2.8 inches, thus $\sigma = 2.8$ .

The interquartile range(IQR) is the difference between the 75th and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = -0.675(2.8)$

$X = 6.91$

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = 0.675(2.8)$

$X = 10.69$

The IQR is:

$IQR = 10.69 - 6.91 = 3.78$

What is the diameter, in inches, of the smallest tree that is an outlier?

The diameter is 1.5IQR above the 75th percentile, thus:

$10.69 + 1.5(3.78) = 16.36$

The diameter of the smallest tree that is an outlier is of 16.36 inches.

A similar problem is given at brainly.com/question/15683591

3 0

2 years ago

Albert is looking up at a tree. he is 25 feet from the vase of tree. the line of sight to the top of the tree is 27 feet.

Studentka2010 [4]

I think 10 but make sure tho

6 0

2 years ago

Read 2 more answers

What will be the index of 2 for its value equal to 1/2

Art [367]

Answer:

the index of 2 is -1

Step-by-step explanation:

$2^{-1}$

= $\frac{1}{2^1}$

= $\frac{1}{2}$

3 0

2 years ago