the histogram to the right displays the height of 122 trees and Mr. Maciag's subdivision. He is very excited about autumn and the leaves changing colors. Height ranged between 108 cm and 345 cm. Let the function "f" map a tree height to the frequency of the column that tree belongs to. What is the value of f(250)? Explain the steps you followed to find the value.​

The scale factor of the dilation is approximately 1.25.

To find the scale factor of the dilation, we need to compare the corresponding side lengths of the original figure and the dilated figure.

Let's consider the two corresponding side lengths:

Original side: EN = 16 units

Dilated side: BN = 20 units

The scale factor (k) of the dilation is the ratio of the corresponding side lengths in the dilated figure to the original figure. It can be calculated using the formula:

Scale factor (k) = Dilated side length / Original side length

Substituting the given values:

k = 20 / 16 ≈ 1.25

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Answer: It takes him 1 minute and 25 seconds

Step-by-step explanation:

5 divided 4

4 times 1 = 4

5-4=1

put the decimal and make 1=10                           Ans.: 1.25

4 times 2 = 8

10-8=2

put the 0 = 20

4 times 5 =20

20-20=100

Answer : You have 23 candy canes

Step - by - step explanation :

23 / 2 = 11 R 1

23 / 3 = 7 R 2

23 / 4 = 5 R 3

Answer:

0.11

Step-by-step explanation:

The events are independent, so:

P(A and B) = P(A) × P(B)

P = 0.5 × 0.22

P = 0.11

Using the concept of binomial probability, the distribution of X which is the number of winning is ;

• X ______ 0 ____1 _____ 2___ 3______ 4

• P(X):_0.0767_0.2762_0.3729_0.2238_0.0503

Given the parmaters :

• Total Number of slots = 38
• Number of reds = 18

The probability of winning,p on a single play :

• p = (red slots / total slots) = 18/38 = 9/19

Number of plays = 4

Number of winnings = X

Using the binomial probability relation :

• P(x = x) = nCx * p^x * q^(n-x)
• q = 1 - 9/19 = 10/19
• n = 4

ForX=0:

P(x = 0) = 4C0 × (9/19)^0 × (10/19)⁴ = 0.0767

ForX=1:

P(x = 0) = 4C1 × (9/19)¹ × (10/19)³ = 0.2762

ForX=2:

P(x = 2) = 4C2 × (9/19)² × (10/19)² = 0.3729

ForX=3:

P(x = 3) = 4C3 × (9/19)³ × (10/19)¹ = 0.2238

For X = 4 :

P(x = 4) = 4C4 × (9/19)⁴ × (10/19)^0 = 0.0503

Therefore, the distribution of X in the roulette spin is :

X ____ 0 ____ 1 _____ 2 ______ 3 ______ 4

P(X):_0.0767_0.2762_0.3729_0.2238 _0.0503

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