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3 years ago

15

Below are the number of flowers the basketball team sold for a fundraiser over a week. What is the mean number of flowers they s

old? Show your work.
18 21 17 24 15

1 answer:

Roman55 [17]3 years ago

4 0

Mean = (18 + 21 + 17 + 24 + 15) divided by 5

Mean = 95 divided by 5

Mean = 19

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John sells hamburgers ($3) and cheeseburgers ($3.50). One afternoon he sells a total of 24 burger for $79. How many of these wer

ankoles [38]

John sold 10 hamburgers and 14 cheeseburgers.

Step-by-step explanation:

Given,

Cost of one hamburger = $3

Cost of one cheeseburger = $3.50

Total burgers sold = 24

Total revenue earned = $79

Let,

Number of hamburgers sold = x

Number of cheeseburgers sold = y

According to given statement;

x+y=24 Eqn 1

3x+3.50y=79 Eqn 2

Multiplying Eqn 1 by 3

$3(x+y=24)\\3x+3y=72\ \ \ Eqn\ 3$

Subtracting Eqn 3 from Eqn 2

$(3x+3.50y)-(3x+3y)=79-72\\3x+3.50y-3x-3y=7\\0.50y=7$

Dividing both sides by 0.50

$\frac{0.50y}{0.50}=\frac{7}{0.50}\\y=14$

Putting y=14 in Eqn 1

$x+14=24\\x=24-14\\x=10$

John sold 10 hamburgers and 14 cheeseburgers.

Keywords: linear equation, elimination method

Learn more about elimination method at:

#LearnwithBrainly

3 0

3 years ago

Mean Value theorem help! BRAINLIEST if you can explain it clearly!!!

Korolek [52]

$\bf f(x)=ln(x-5)\qquad [6,8]\qquad \cfrac{df}{dx}=\cfrac{1}{x-5}\\\\
-----------------------------\\\\
\textit{mean value theorem}\qquad f'(c)=\cfrac{f(b)-f(a)}{b-a}\\\\
-----------------------------\\\\
f'(c)=\cfrac{1}{c-5}\qquad thus\quad \cfrac{1}{c-5}=\cfrac{f(8)-f(6)}{8-6}
\\\\\\
\cfrac{1}{c-5}=\cfrac{ln(3)-ln(1)}{2}\implies \cfrac{1}{c-5}=\cfrac{ln(3)-0}{2}
\\\\\\
\cfrac{2}{ln(3)}=c-5\implies \boxed{\cfrac{2}{ln(3)}+5=c}$

7 0

3 years ago

Why would you use the measurement 8 feet 4 inches instead of 7 feet 16 inches?.

lidiya [134]

8 feet 4 inches is more precise and easier to use.

It would be weird to say 7 feet 16 inches, because there's 12 inches in a foot and your already using that measurement.

Hope this is helpful:)

3 0

3 years ago

At the start of the year, 15 chameleons were introduced into a zoo. The population of chameleons is expected to grow at a rate o

bearhunter [10]

Answer:

option-B

Step-by-step explanation:

We are given

At the start of the year, 15 chameleons were introduced into a zoo

so, $P_0=15$

The population of chameleons is expected to grow at a rate of 41.42% every year

so, r=0.4142

and x represents the number of years since the chameleons were introduced into the zoo

now, we can set equation to find total population

and we get

$P(x)=P_0(1+r)^x$

now, we can plug values

$P(x)=15(1+0.4142)^x$

$P(x)=15(1.4142)^x$

Average rate of change between 2 years and 4 years:

we can use formula

$A_1=\frac{P(4)-P(2)}{4-2}$

now, we can plug values

$A_1=\frac{15(1.4142)^{4}-15(1.4142)^{2}}{4-2}$

$A_1=14.99914$

Average rate of change between 4 years and 6 years:

we can use formula

$A_2=\frac{P(6)-P(4)}{6-4}$

now, we can plug values

$A_2=\frac{15(1.4142)^{6}-15(1.4142)^{4}}{6-4}$

$A_2=29.99770$

Average rate of change between 6 years and 8 years:

we can use formula

$A_3=\frac{P(8)-P(6)}{8-6}$

now, we can plug values

$A_3=\frac{15(1.4142)^{8}-15(1.4142)^{6}}{8-6}$

$A_3=59.99425$

now, we will check each options

option-A:

we can see that

$A_3-A_2=30$

$A_3-A_2=30$

So, this is FALSE

option-B:

$A_1=\frac{1}{2}A_2$

So, this is TRUE

option-C:

This is FALSE

option-D:

we got

$A_1=\frac{1}{2}A_2$

so, this is FALSE

5 0

3 years ago

7. The football team has three games remaining on its schedule. At the beginning of each

docker41 [41]

Answer:

B

Step-by-step explanation:

5 0

3 years ago