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3 years ago

11

Harold measures the heights in inches of the new plants that are growing in his greenhouse. He records the lengths shown below a

nd wants to plot them on a line plot. 19.5 21.5 18.0 19.5 19.25 19.5 21.0 18.25 20.5 18.75 19.0 19.0 19.25 22.0 Use the drop-down menus to complete the following sentences about the line plot. The line plot has dots plotted on it. The least value plotted on the line plot is . The greatest value plotted on the line plot is . The height of inches occurs the most often.

1 answer:

viva [34]3 years ago

8 0

Answer:

The least value plotted on the line plot is 18.0

The greatest value plotted on the line plot is 22.0

The height of inches occurs the most often is 19.5 with 3 times

Step-by-step explanation:

A line plot is a graph that shows frequency (the number of times the data value occur) of data along a number line.

The line plot has dots plotted on it over each value of the number line.

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If Jace had 77 blue pens,23 red pens 15 pink pens and 4 purple pens what is the part to whole ratio for red pens to all of the p

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Answer:

23:119

Step-by-step explanation:

all pens:

77 blue + 23 red + 15 pink + 4 purple

77 + 23 + 15 + 4 = 119

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23 red / 119 total

23/119

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jolli1 [7]

Answer:

Part 1) $r=3.5\%$

Part 2) $t=6\ years$

Part 3) $21.9\ ft$

Step-by-step explanation:

Part 1) we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

Let

x -----> amount of money to be invested

2x ---> Final Investment Value

we have

$t=20\ years\\ P=\$x\\A=\$2x\\r=?$

substitute in the formula above

$2x=x(e)^{20r}$

$2=(e)^{20r}$

Apply ln both sides

$ln(2)=ln[(e)^{20r}]$

$ln(2)=(20r)ln(e)$

$ln(2)=(20r)$

$0.693=(20r)$

$r=0.693/20$

$r=0.035$

$r=3.5\%$

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

Let

x -----> amount of money to be invested

2x ---> Final Investment Value

we have

$t=?\ years\\ P=\$x\\A=\$2x\\r=0.126$

substitute in the formula above

$2x=x(e)^{0.126t}$

$2=(e)^{0.126t}$

Apply ln both sides

$ln(2)=ln[(e)^{0.126t}]$

$ln(2)=(0.126t)ln(e)$

$ln(2)=(0.126t)$

$0.693=(0.126t)$

$t=0.693/0.126$

$t=5.5=6\ years$

Part 3) we have

$f(x)=18.4+2.5ln(x+1)$

where

f(x) -----> is the long jump record in feet

x -----> the number of years since the school was opened

so

For x=3 years

substitute

$f(3)=18.4+2.5ln(3+1)$

$f(3)=18.4+2.5ln(4)$

$f(3)=18.4+2.5(1.386)$

$f(3)=21.9\ ft$

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