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3 years ago

Mr. Evans displays the test scores of his first period class in the box plot below.

Mathematics

2 answers:

aleksandr82 [10.1K]3 years ago

8 0

Answer:

Step-by-step explanation:

Median is the middle entry when arranged in ascending or descending order.Q1 can be mentioned as the median of first entry to median, when arranged inascending order. Similarly Q3 is the median of the entries from median to end

Q1=78

IQR=Q3-Q1=5

Q3=78+5=83

Hence answer is 83

Illusion [34]3 years ago

7 0

The IQR of the first period is 15. The IQR of the third period is 1/3 that, 5*(1/3) is equal to 5.

For the third period, Q1 is 78. The expression for Q3 is Q1+IQR

Therefore, 78+5 = 83

Q3 of the third period is 83

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PolarNik [594]

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3 years ago

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Serhud [2]

Answer:

C. 14xy - 6y²

Step-by-step explanation:

The distributive property will essentially result in the sum of the product of the "outside term" with each of the "inside terms".

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3 years ago

3 + 3 + 3 +<br> + 4 + 4 =

Gennadij [26K]

Answer:

3+3+3+3=12

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3 years ago

Read 2 more answers

A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft and on the other three s

makvit [3.9K]

Answer:

The dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

Step-by-step explanation:

Let the length of garden be x

Let the breadth of garden be y

Area of Rectangular garden = $Length \times Breadth = xy$

We are given that the area of the garden is 122 square feet

So, $xy=122$ ---A

A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft

So, cost of brick along length x = 20 x

On the other three sides by a metal fence costing $10/ft.

So, Other three side s = x+2y

So, cost of brick along the other three sides= 10(x+2y)

So, Total cost = 20x+10(x+2y)=20x+10x+20y=30x+20y

Total cost = 30x+20y

Substitute the value of y from A

Total cost = $30x+20(\frac{122}{x})$

Total cost = $\frac{2440}{x}+30x$

Now take the derivative to minimize the cost

$f(x)=\frac{2440}{x}+30x$

$f'(x)=-\frac{2440}{x^2}+30$

Equate it equal to 0

$0=-\frac{2440}{x^2}+30$

$\frac{2440}{x^2}=30$

$\sqrt{\frac{2440}{30}}=x$

$9.018 =x$

Now check whether it is minimum or not

take second derivative

$f'(x)=-\frac{2440}{x^2}+30$

$f''(x)=-(-2)\frac{2440}{x^3}$

Substitute the value of x

$f''(x)=-(-2)\frac{2440}{(9.018)^3}$

$f''(x)=6.6540$

Since it is positive ,So the x is minimum

Now find y

Substitute the value of x in A

$(9.018)y=122$

$y=\frac{122}{9.018}$

$y=13.528$

Hence the dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

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3 years ago

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Mrrafil [7]

Answer:

15,103.05

Step-by-step explanation:

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3 years ago