3m^n+12mn^2 with an ex

A tree farmer Planet 3 types of trees on 22 acres of land .he planted 48 trees per acre.what was the total number of trees the f

Alja [10]

Answer:

The total number of trees the farmer planted = 1056 trees

Step-by-step explanation:

It is given that,

A tree farmer Planet 3 types of trees on 22 acres of land and,

he planted 48 trees per acre

<u>To find total number of plants</u>

There are 22 acres of land.

In one acre he planted 48 trees.

Therefore total number of plants in 22 acers = 22 * 48 = 1056 trees

5 0

2 years ago

Three boxes are stacked one on top of the other. One box is 5 feet 10 inches tall , one is 4 feet 2 inches tall and one is 2 fee

katovenus [111]

The boxes would be 12 feet and 3 inches tall

6 0

3 years ago

The frequency distributions of two data sets are shown in the dot plots below.

fredd [130]

Answer:

1st, 4th and 6th statements are true.

Step-by-step explanation:

We have been given two dot-plots and we are asked to select all the true statements about given dot plots.

Let us write all the data points of both dot plots.

Data set 1:

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 7.

Data set 2:

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4.

1. The mean of data set 1 is greater than the mean of data set 2.

Let us find mean of both data sets.

$\text{Mean of data set 1}=\frac{1+1+1+1+1+2+2+2+2+2+2+2+2+3+3+3+4+4+7}{19}$

$\text{Mean of data set 1}=\frac{45}{19}$

$\text{Mean of data set 1}=2.3684\approx 2.37$

$\text{Mean of data set 2}=\frac{1+1+1+1+1+2+2+2+2+2+2+2+2+3+3+3+4+4}{18}$

$\text{Mean of data set 2}=\frac{38}{18}$

$\text{Mean of data set 2}=2.11$

We can see that mean of data set 1 is greater than mean of data set 2, therefore, 1st statement is true.

2. The mean of data set 1 is equal to the mean of data set 2.

Since mean of data set 1 is greater than mean of data set 2, therefore, 2nd statement is false.

3. The median of data set 1 is greater than the median of data set 2.

Since data set 1 has 19 data points, so median of data set 1 will be value of 10th data set, that is 2.

Our data set 2 has 18 data points, so median of data set 2 will be the average of 9th and 10th data points.

$\text{Median of data set 2}=\frac{2+2}{2}$

$\text{Median of data set 2}=\frac{4}{2}=2$

Since median of data set 2 is also 2, therefore, 3rd statement is not true.

4. The median of data set 1 is equal to the median of data set 2.

Since both data set's median is 2, therefore, 4th statement is true.

5. The standard deviation of data set 1 is smaller than the standard deviation of data set 2.

Using online SD calculator we get SD of data set 1 is 1.422 and SD of data set 2 is 0.936. Since 1.422 is greater than 0.936, therefore, 5th statement is false.

6. The data sets have the same interquartile range.

$Q_1\text{ of data set 1}=1$

$Q_3\text{ of data set 1}=3$

$\text{IQR of data set 1}=3-1=2$

$Q_1\text{ of data set 2}=1$

$Q_3\text{ of data set 2}=3$

$\text{IQR of data set 2}=3-1=2$

Since both data set's IQR is 2, therefore, 6th statement is true.

3 0

3 years ago

Read 2 more answers

Evaluate 2xcubed if x= 3/4

Monica [59]

9514 1404 393

Answer:

27/32

Step-by-step explanation:

2x³ for x = 3/4 is ...

2(3/4)³ = 2·3³/4³ = 2·27/64 = 27/32

6 0

2 years ago

I NEED A ANSWER FAST PLEASE HELP

NikAS [45]

Answer:

Elevation in boiling point is mathematically expressed as

ΔTb = Kb X m

where Kb = boiling point elevation constant = 2.53°c/m (for benzene)

m = molality of solution = 3.47 m (given)

∴ ΔTb = 2.53 X 3.47

= 8.779 oC

But, boiling point of benzene = 80.1 oC

∴ Boiling point of solution = 88.879 oC

Now, Depression in freezing point = ΔTf = Kf X m

where, Kf = freezing point depression constant = 5.12°c/m (for benzene)

∴ ΔTf = 5.12 X 3.47

= 17.766 oC

But freezing point of benzene = 5.5 oC

∴ Freezing point of solution = -12.266 oC

3 0

2 years ago