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artcher [175]
3 years ago
14

A gardener will use up to 250 square feet for planting flowers and vegetables. He wants the area used for flowers to be at least

three times the area used for vegetables. Let x denote the area (in square feet) used for flowers. Let y denote the area (in square feet) used for vegetables. Shade the region corresponding to all values of x and that y satisfy these requirements.
Mathematics
1 answer:
anyanavicka [17]3 years ago
8 0
The answer 
<span>Let x denote the area (in square feet) used for flowers.
</span><span>Let y denote the area (in square feet) used for vegetables.

we have    x + y =250
                 x<= 3y

</span>x + y =250
x - 3y <= 0

graph this system and <span>shade the region corresponding to all values of x and that y satisfy these requirements.</span>
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Dr drake is thinking about retirement and decides to sail around the world once he retires. He buys a sailboat for 125000. He bo
Akimi4 [234]

Answer:

$46875.

Step-by-step explanation:

We have been given that Dr. Drake buys a sailboat for 125000. He borrows the money at an APR of 7.5% for five years.

We will find total interest after 5 years by using simple interest formula.

A=P(1+rT), where,

A= Amount after T years.

P= Principal amount.

r= The annual interest rate (in decimal form).

T= Time in years.

First of all we will convert our given interest rate from percent to decimal form.

7.5\text{ Percent}=\frac{7.5}{100} =0.075

Now let us substitute our given values in above formula.

A=125,000(1+(0.075*5))

A=125,000(1+0.375)

A=125,000(1.375)

A=171875

To find the amount of total interest we will subtract principal amount from 171875.

\text{Total amount of interest}=171875-125000

\text{Total amount of interest}=46875  

Therefore, the total interest after 5 years will be $46875.

8 0
4 years ago
The table shows the hourly cookie sales by students in grades 7 and 8 at the school's annual bake sale. Grade 7 Grade 8 20 21 15
Naddika [18.5K]

Answer:

1. The interquartile range for the grade 7 data is 6.

2. The interquartile range for the grade 8 data is 6.

3. The difference of the medians of the two data sets is 2.

4. The difference is about 1/3 times the interquartile range of either data set.

Step-by-step explanation:

The hourly cookie sales by students in grades 7 and 8 at the school's annual bake sale is given by the following table.

Grade 7       Grade 8

  20                  21

  15                  29

  30                  14

  24                  19

   18           24

   21                 25

The data set for grade 7 is

20, 15, 30, 24, 18, 21

Arrange the data in ascending order.

15, 18, 20, 21, 24, 30

Divide the data in four equal parts.

(15), 18, (20), (21), 24,( 30)

Q_1=18, Median=\frac{20+21}{2}=20.5, Q_3=24

The interquartile range for the grade 7 data is

IQR=Q_3-Q_1=24-18=6

Therefore the interquartile range for the grade 7 data is 6.

The data set for grade 8 is

21, 29, 14, 19, 24, 25

Arrange the data in ascending order.

14, 19, 21, 24, 25, 29

Divide the data in four equal parts.

(14), 19, (21), (24), 25, (29)

Q_1=19, Median=\frac{21+24}{2}=22.5, Q_3=25

The interquartile range for the grade 8 data is

IQR=Q_3-Q_1=25-19=6

Therefore the interquartile range for the grade 8 data is 6.

The difference of the medians of the two data sets is

D=22.5-20.5=2

Therefore the difference of the medians of the two data sets is 2.

Let the difference is about x times the interquartile range of either data set.

The IQR of each data is 6.

D=x(IQR)

2=x(6)

\frac{2}{6}=x

\frac{1}{3}=x

Therefore the difference is about 1/3 times the interquartile range of either data set.

8 0
3 years ago
Help me ASAP it’s a math test
expeople1 [14]

Answer:

I would say the last one, bc b Is negative correct?

4 0
3 years ago
Really do need help on this
melamori03 [73]
The answer would be d blue  because blue  has less  colors than the rest of them.
7 0
3 years ago
Find the value and the compound interest if £4000 is invested for 2 years at 36% p.a.
Andrei [34K]
  • Principal=P=4000
  • Time=T=2years
  • Rate of interest =R=36

Interest be I

\\ \sf\longmapsto  I=\dfrac{PRT}{100}

\\ \sf\longmapsto  I=\dfrac{4000(2)(36)}{100}

\\ \sf\longmapsto  I=\dfrac{288000}{100}

\\ \sf\longmapsto  I=2880

Now

Final value=4000+2880=6880

4 0
3 years ago
Read 2 more answers
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