What is the equation for synthetic division

You come home from Brian’s Orchard with a big brown bag of apples: 23 Granny Smiths, 14 Honey Crisp and 31 Red Delicious. What i

garik1379 [7]

Answer:

a. Probability of Pulling one of each = 0.03175

b. Probability of Pulling 4 Honey Crisp = 0.001797

Probability of 2 G.Smith = 0.1144

Probability of 1 Red Delicious = 0.4559

Step-by-step explanation:

Given

Granny Smiths = 23

Honey Crisp = 14

Red Delicious = 31

Required

- Probability of Pulling out one of each

- Probability of Pulling out 4 Honey Crisp; 2 Granny Smiths; 1 Red Delicious

First, the total number of apple needs to be calculated.

Total = Granny Smiths + Honey Crisp + Red Delicious

Total = 23 + 14 + 31

Total = 68

Probability of Pulling 1 of each

= P(Granny Smiths) and P(Henry Ford) and P(Red Delicious)

- Granny Smiths;

This is calculated by dividing number of Granny Smiths apples by total number of apples.

Probability = 23/68

Similarly,

Probability of Pulling Honey Crisp= Number of Honey Crisp divided by total

Probability = 14/68

Probability = 7/34

Probability of Pulling Red Delicious = Number of Red Delicious divided by total

Probability = 31/68

So, Probability of Pulling 1 of each = 23/68 * 7/34 * 31/68

Probability = 4991/157216

Probability = 0.03175

Probability of Pulling out 4 Honey Crisp;

= P(Honey) * P(Honey) * P(Honey) * P(Honey)

= (P(Honey))⁴

= (7/34)⁴

= 2401/1336336

= 0.001797

Probability of Pulling 2 Granny Smiths;

= P(Granny) * P(Granny)

= (P(Granny))²

= (23/68)²

= 529/4624

= 0.1144

Probability of 1 Red Delicious

= number of red delicious divided by total

= 31/68

= 0.4559

3 0

3 years ago

The answer to B and C

ArbitrLikvidat [17]

Assuming that the triangles are congruent, you can just look at the parts on each triangle to find the parts on the other triangle. The answers are yes, yes, and no. The triangles are obviously not drawn to scale, so just remember to never rely on it being so.

3 0

3 years ago

A skier has decided that on each trip down a slope, she will do 3 more jumps than before. On her first trip she did 5 jumps. Der

taurus [48]

Since we are already given the amount of jumps from the first trial, and how much it should be increased by on each succeeding trial, we can already solve for the amount of jumps from the first through tenth trials. Starting from 5 and adding 3 each time, we get: 5 8 (11) 14 17 20 23 26 29 32, with 11 being the third trial.

Having been provided 2 different sigma notations, which I assume are choices to the question, we can substitute the initial value to see if it does match the result of the 3rd trial which we obtained by manual adding.

Let us try it below:

Sigma notation 1:

10
 Σ (2i + 3)
i = 3

@ i = 3

2(3) + 3
12

The first sigma notation does not have the same result, so we move on to the next.

10
 Σ (3i + 2)
i = 3

When i = 3; 3(3) + 2 = 11. (OK)

Since the 3rd trial is a match, we test it with the other values for the 4th through 10th trials.

When i = 4; 3(4) + 2 = 14. (OK)
When i = 5; 3(5) + 2 = 17. (OK)
When i = 6; 3(6) + 2 = 20. (OK)
When i = 7; 3(7) + 2 = 23. (OK)
When i = 8; 3(8) + 2 = 26. (OK)
When i = 9; 3(9) + 2 = 29. (OK)
When i = 10; 3(10) + 2 = 32. (OK)

Adding the results from her 3rd through 10th trials: 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 = 172.

Therefore, the total jumps she had made from her third to tenth trips is 172.

3 0

3 years ago

Two trains start from Fort Worth traveling at the same speed. The trip for Train W takes 20 hours, while the trip for Train Z ta

MrMuchimi

Answer:

The equation that best models this situation is :

b. $\frac{x}{20}=\frac{x+400}{25}$