Kenji walks 44 feet in 10 seconds. At this rate, how many miles does Kenji walk in an hour? show your work. (1 mile = 5,280 feet

The provided values are

$85 $240 $120 $60 $315 $190 $145 $105 $285 $260 $95 $215 $235 $175 $155

To start, we will arrange these in ascending order

resulting in

$60 $85 $95  $105  $120  $145 $155 $175 $190 $215 $235 $240 $260 $285 $315

The total count is 15

the position of 105 is 4

Now, we can calculate the percentile

Thus, the percentile for the bid of $105 is 26.667%...........Response

7 hours 12 minutes
6 hours 46 minutes
6 hours 53 minutes
--------------------------sum
Total: 19 hours 111 minutes, which converts to (19 × 60) + 111 = 1140 + 111 = 1251 minutes

Calculate average time by dividing by 3: 1251 / 3 = 417 minutes, equivalent to 6 hours 57 minutes, which is the average delivery duration.

Answer:

The accurate assertions include:

A. m∠6 = 55°

C. m∠1 + m∠4 = 250°

D. m∠1 + m∠6 = m∠7 + m∠4

Step-by-step clarification:

The provided information is as follows:

m∠7 = 55°

The angles formed by the transversal and the upper horizontal parallel line are (starting from the top left and moving in clockwise direction) 1, 2, 4, 3

Likewise, the angles from the transversal and the lower horizontal parallel line are (starting from the top left and going clockwise) 5, 6, 8, 7

Consequently, we have;

m∠7 ≅ m∠6 (Vertically opposite angles are equal)

Thus, m∠6 = m∠7 = 55°

m∠6 = 55°, which aligns with option A.

m∠5 + m∠6 = 180° (The sum of angles on a straight line)

So, m∠5 = 180° - m∠6 = 180° - 55° = 125°

m∠5 = 125°

m∠1 ≅ m∠5 (Corresponding angles)

Thus, m∠1 = m∠5 = 125°

m∠1 ≅ m∠4 (Vertically opposite angles)

Therefore, m∠1 = m∠4 = 125°

Thus, m∠1 + m∠4 = 125° + 125° = 250°

m∠1 + m∠4 = 250°, which corresponds to option C.

m∠1 ≅ m∠4 (Vertically opposite angles)

m∠6 ≅ m∠7 (Vertically opposite angles)

Thus, m∠1 + m∠6 = m∠7 + m∠4 (Transitive property), which matches option D.

Answer: The mean and variance of Y are $0.25 and $6.19 respectively.

Step-by-step explanation:

The scenario is as follows: You and a friend participate in a game involving tossing a fair coin.

The sample space for tossing two coins is {TT, HT, TH, HH}

Let Y represent the earnings from one round of the game.

If both faces are heads, you win $1; therefore, P(Y=1)=P(TT)=

You win $6 if both faces are heads, so P(Y=6)=P(HH)=

If the faces do not match, you lose $3 which means P(Y=1)=P(TH, HT)=

To find the expected value to win: E(Y)=

Thus, the mean of Y: E(Y)= $0.25

Variance =

Therefore, variance of Y = $ 6.19

Starting with the equation 4(3b + 2)² = 64, if we divide both sides by 4 we obtain (3b + 2)² = 16. By taking the square root of both sides, we derive two cases: (3b + 2) = 4 and (3b + 2) = -4. Solving each equation for b yields: 3b = 2 or 3b = -6, leading to b values of 2/3 and -2. Ultimately, the results specify that b = 2/3 and b = -2.