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Ilia_Sergeevich [38]
2 years ago
9

If a 40 foot tree casts a 12 foot shadow find the length of the shadow cast by a 24 foot tree

Mathematics
1 answer:
Akimi4 [234]2 years ago
3 0

40 foot tree = 12 foot shadow
24 foot tree = x foot shadow

(Cross Multiply)

40 × x = 12 × 24
x = 288 / 40
x = 7.2 foot

24 foot tree casts 7.2 foot shadow
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(5,6) and (−5,12) what is the rate of change??
Veseljchak [2.6K]

Answer:

- 0.6

Step-by-step explanation:

Start point: (5,6)

End point: (-5,12)

rate of change = (12-6) / (-5-5) = 6/-10 = - 3/5 = - 0.6

4 0
2 years ago
Read 2 more answers
You have to make 1000 buttons.the diameter of the button is 9cm but you have to find it’s area.its $15/m squared and you have to
Semmy [17]

Answer:

$95.38

Step-by-step explanation:

step 1

Find the area of one button

The area is equal to

A=\pi r^{2}

we have

r=9/2=4.5\ cm ---> the radius is half the diameter

Convert to meters

r=4.5\ cm=4.5/100=0.045\ m

assume

\pi =3.14

substitute

A=(3.14)(0.045)^{2}\\A=0.0063585\ m^2

step 2

Find the area of 1,00 buttons

Multiply by 1,000

A=0.0063585(1,000)=6.3585\ m^2

step 3

Find the cost

Multiply $15 per square meter by the total area of 1,000 buttons

(15)6.3585=\$95.38

3 0
3 years ago
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7 0
2 years ago
PLEASE HELP ME
denis23 [38]

Answer:

(1) The possible outcomes are: X = {0, 1, 2, 3}.

(2) The number of times should Hartley spin a difference of 1 is 36.

(3) The number of times should Hartley spin a difference of 0 is 24.

Step-by-step explanation:

The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.

The total number of spins for each of the spinner is, <em>n</em> = 96.

(1)

The sample space of spinning both the spinners together are:

S = {(1, 1), (1, 2), (1, 3), (1, 4)

      (2, 1), (2, 2), (2, 3), (2, 4)

      (3, 1), (3, 2), (3, 3), (3, 4)

      (4, 1), (4, 2), (4, 3), (4, 4)}

Total = 16.

The possible outcomes are:

X = {0, 1, 2, 3}.

(2)

The sample space with the difference 1 are:

S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

n (S₁) = 6

The probability of the difference 1 is:

P(\text{Diff}=1)=\frac{n(S_{1})}{N}=\frac{6}{16}=\frac{3}{8}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 1 is:

E(\text{Diff}=1)=P(\text{Diff}=1)\times n\\\\=\frac{3}{8}\times 96\\\\=36

Thus, the number of times should Hartley spin a difference of 1 is 36.

(3)

The sample space with the difference 0 are:

S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}

n (S₂) = 4

The probability of the difference 0 is:

P(\text{Diff}=0)=\frac{n(S_{2})}{N}=\frac{4}{16}=\frac{1}{4}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 0 is:

E(\text{Diff}=0)=P(\text{Diff}=0)\times n\\\\=\frac{1}{4}\times 96\\\\=24

Thus, the number of times should Hartley spin a difference of 0 is 24.

4 0
2 years ago
Help please? And show work?
In-s [12.5K]
Answer the answer is 25




Explanation:
6 0
3 years ago
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