(3.14) please help me as soon as possible ....

Based on the cumulative frequency histogram, determine the number of swimmers who swam between 200 and 249 yards. Determine the

Lostsunrise [7]

Answer:

$(a)\ 200 \to 249 =3$

$(b)\ 150 \to 199 = 0$

$(c)\ Total = 20$

Step-by-step explanation:

Given

See attachment for cumulative frequency histogram

Solving (a): Swimmers between 200yd and 249yd

To do this, we simply read the data from the 0 mark

From the histogram, we have:

$0 \to 249 = 15$

and

$0 \to 199 = 12$

So:

$200 \to 249 = 0 \to 249 - 0 \to 199$

This gives:

$200 \to 249 = 15-12$

$200 \to 249 =3$

Solving (b): Swimmers between 150yd and 199yd

To do this, we simply read the data from the 0 mark

From the histogram, we have:

$0 \to 149 = 12$

and

$0 \to 199 = 12$

So:

$150 \to 199 = 0 \to 199 - 0 \to 149$

This gives:

$150 \to 199 = 12-12$

$150 \to 199 = 0$

Solving (c): Total swimmers

To do this, we simply read the longest bar of the histogram

$Longest = 20$

Hence:

$Total = 20$

4 0

2 years ago

Can someone explain step by step? I’m confused. (Mark for brainliest!)

NikAS [45]

Answer:

190

Step-by-step explanation:

57.5 x 2=115 so that is the arc for CE

then you add that to 55 to get 170

a circle is 360 so then you subtract 360 for 170

6 0

3 years ago

Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible

Law Incorporation [45]

Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

$pq = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

$or \: \sqrt{( {5 + 3)}^{2} + ( {2 - 2)}^{2} }$

By simplifying it we get,

$8$

Now,

again finding the distance between PR,

$pr = \sqrt{( {x2 - x1}^{2} + ( {y2 - y1)}^{2} }$

or,

$\sqrt{( {x + 3)}^{2} + ( {y - 2)}^{2} }$

By simplifying it we get,

$= \sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 }$

now, finding the distance of QR,

$qr = \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2} }$

or, by simplification we get,

$\sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

now, equating PR and QR,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13} = \sqrt{ {x}^{2} + {y}^{2} - 10x - 4y + 29 }$

we cancelled the root ,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = {x}^{2} + {y}^{2} -10x - 4y + 29$

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

$\sqrt{ {x}^{2} + {y}^{2} + 6x - 4y + 13 } = 8}$

cancel the roots,

${x}^{2} + {y}^{2} + 6x - 4y + 13 = 8$

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

Hope it helps...

3 0

3 years ago

What is the average rate of change of f(x), represented by the graph, over the interval [-2, 2]?

Molodets [167]

To get the average rate of change (ARC) of f(x) over [x1, x2], we use the formula:
ARC = ( f(x2) - f(x2) ) / (x2 - x1)

From the graph
f(2) = 4
f(-2) = 4

Plugging in the values into the formula:
ARC = (4 - 4) / (2 - (-2) )
ARC = 0

The points connecting (-2,4) amd (2,4) is a horizontal line that is the rate of change is 0.

5 0

3 years ago

Read 2 more answers

A chocolate shop sold 1 1/4 ounces of white chocolate. It sold 8 1/4 times as much milk chocolate as white chocolate. How many o

rjkz [21]

Answer:

10.3125 (NOT ROUNDED)

Step-by-step explanation:

1. I changed it to decimals. 1/4 can easily be changed into a decimal.

2. I set 1/4 = x/100. Then cross multiplied. That would give you 4x = 100.

3. Now you would simplify that. x would be 25. So its 25/100 wich then turns to .25.

4. Now you are going to replace them so it would be 1.25 and 8.25.

5. To find out how much milk chocolate there was you would multiply 1.25 and 8.25.

6. That's your final answer

6 0

3 years ago