MATHEMATICS COLLEGE

Find the slope using the formula.
(4, 19) and (2, 11)

Answers

Answer 1

Answer:

Answer:

$m=4$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Slope Formula: $m=(y_2-y_1)/(x_2-x_1)$

Step-by-step explanation:

Step 1: Define

(4, 19)

(2, 11)

Step 2: Find slope m

Substitute: $m=(11-19)/(2-4)$
Subtract: $m=(-8)/(-2)$
Divide: $m=4$

Related Questions

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Karly is choosing between two health clubs. Yoga Studio A charges a membership fee of $22.00 and $24.50 per month. Yoga Studio B charges a membership fee of $47.00 and $18.25 per month. At how many months will they cost the same? *
Miguel has a jar full of dimes and quarters. There are a total of 115 coins with a total value of$17.05. The system of equations that represents this situation is as follows: d + q =115 and10d + 25 = 1705 How many MORE dimes than quarters are in Miguel's jar?
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
arlene sells computers and tablets.she earns an 8% commision on every dollar of sales that she makes.In one month she earned a total of 2,560 in commisions.write an equation for Arlene sales

Please answer correctly!

Graphing the equation y-3=1/3(x+4)

Thank you.

Answers

Answer:

X intercept (-13,0)

y intercept (0,13/3)

Step-by-step explanation:

slope intercept form

y=1/3x+13/3

Find the product of (5.2 · 10^-6) and (8 · 10^3).A) 416 · 106-4
B) 4.16 · 10^-2
C) 41.6 · 10^-3
D) 0.416 · 10^-1

Answers

Answer: B) $4.16\cdot10^(-2)$

Step-by-step explanation:

The given product : $(5.2\cdot10^(-6))\cdot (8\cdot10^3)$

First open parenthesis :

$5.2\cdot10^(-6)\cdot 8\cdot10^3$

Write decimal values together and power of 10s together.

$5.2\cdot 8\cdot10^(-6)\cdot10^3$

Using Law of exponent : $a^m\cdot a^n= a^(m+n)$

The above expression becomes.

$41.6\cdot10^(-6+3)=41.6*10^(-3)$

In scientific notation, the decimal must be placed after one digit (from left).

$41.6*10^(-3)=4.16*10*10^(-3)\n\n=4.16\cdot10^(-3+1)\n\n=4.16\cdot10^(-2)$

Hence, the correct answer is B) $4.16\cdot10^(-2)$ .

$\displaystyle\n(5.2\cdot10^(-6))*(8\cdot10^3)=\n\n=\underbrace{5.2*8}_(41.6)*\underbrace{10^(-6)*10^(3)}_(10^(-6+3))=\n\n=41.6*10^(-6+3)=\boxed{\bf41.6*10^(-3)}\n\n\texttt{Correct answer:}~~\boxed{\bf C)}$

5. Taking a cruise is a costly discretionary expense. In a recent year, the top fivecruise lines in the world had this many passengers:
4,133,000 2,369,000 1,295,000 928,000 679,000
Round your answers to the nearest integer.
a. The computations will be easier to work if you view this problem in terms
of thousands of passengers. Represent each number in terms of thousands
of passengers.
b. What is the mean number of passengers for these five cruise lines? (Give
the full number.)
c. What is the range? (Give the full number.)
d. What is the standard deviation? (Give the full number.)

Answers

Final answer:

The numbers in thousands are: 4133, 2369, 1295, 928, 679. The mean is 1881 thousand passengers, the range is 3454 thousand passengers, and the standard deviation is approximately 1218 thousand passengers.

Explanation:

Part A: To represent each number in terms of thousands of passengers, we simplify them as follows: 4,133,000 is 4,133 thousands of passengers, 2,369,000 is 2,369 thousands, 1,295,000 is 1,295 thousands, 928,000 is 928 thousands, and 679,000 is 679 thousands.

Part B: To calculate the mean number of passengers, you would add up all the passenger numbers and then divide by the number of values (5 in this case). This adds up to 9,404,000 passengers or, in terms of thousands, 9,404. Divided by 5, this gives a mean of 1,880,800 passengers, or 1,881 thousands of passengers.

Part C: The range is calculated by subtracting the smallest number from the largest. In this case, that would be 4,133,000 - 679,000 = 3,454,000, which is also 3,454 thousands of passengers.

Part D: Calculating the standard deviation involves multiple steps. First, for each value, subtract the mean and square this result. Then, calculate the mean of these squared differences. Finally, take the square root of this mean. Doing so, the standard deviation is approximate 1,217,982 passengers, or 1,218 thousands of passengers.

Learn more about Statistics here:

brainly.com/question/31538429

#SPJ12

Convert 3,A5D Base 16 to Base 10

Answers

Answer:

14941.

Step-by-step explanation:

In base 16 we have that :

A=10, B=11, C=12, D=13, E=14, F=15 and the process of change is:

3: $3*16^(0)= 3$

A5D= $10*16^(2)+5*16^(1)+13*16^(0)= 2560+80+13=2653.$

3A5D = $3*16^(3)+10*16^(2)+5*16^(1)+13*16^(0)= 12288+2560+80+13=14941.$

Given line AB with point P above, Grace was asked to construct a line parallel to line AB. She used her compass and straight edge to draw line PQ parallel to line AB using a transversal line PT. Which statement is true? A) AT = QP B) PR = TB C) ∠PTB = ∠PTA D) ∠SPR = ∠PTB

Answers

The answer is D i took the test

D

The transversal line PT does not bisect either line AB nor QR, so there are no equal line segments. Angles PTB and PTA are complementary, not equal in measure. The true statement is ∠SPR = ∠PTB

The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and standard deviation 129 chips.(a) What is the probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive? (b) What is the probability that a randomly selected bag contains fewer than 1025 chocolate chips? (c) What proportion of bags contains more than 1200 chocolate chips? (d) What is the percentile rank of a bag that contains 1425 chocolate chips?

Answers

Using the normal distribution, it is found that:

a) There is a 0.947 = 94.7% probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive.

b) There is a 0.0392 = 3.92% probability that a randomly selected bag contains fewer than 1025 chocolate chips.

c) 0.3446 = 34.46% of bags contains more than 1200 chocolate chips.

d) The bag is in the 91st percentile.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = (X - \mu)/(\sigma)$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

Mean of 1252 chips, thus $\mu = 1252$ .
Standard deviation of 129 chips, thus $\sigma = 129$ .

Item a:

The probability is the p-value of Z when X = 1500 subtracted by the p-value of Z when X = 1000, thus:

X = 1500:

$Z = (X - \mu)/(\sigma)$

$Z = (1500 - 1252)/(129)$

$Z = 1.92$

$Z = 1.92$ has a p-value of 0.9726.

X = 1000:

$Z = (X - \mu)/(\sigma)$

$Z = (1000 - 1252)/(129)$

$Z = -1.95$

$Z = -1.95$ has a p-value of 0.0256.

Then, 0.9726 - 0.0256 = 0.947

There is a 0.947 = 94.7% probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive.

Item b:

This probability is the p-value of Z when X = 1025, thus:

$Z = (X - \mu)/(\sigma)$

$Z = (1025 - 1252)/(129)$

$Z = -1.76$

$Z = -1.76$ has a p-value of 0.0392.

There is a 0.0392 = 3.92% probability that a randomly selected bag contains fewer than 1025 chocolate chips.

Item c:

This proportion is 1 subtracted by the p-value of Z when X = 1200, thus:

$Z = (X - \mu)/(\sigma)$

$Z = (1200 - 1252)/(129)$

$Z = -0.4$

$Z = -0.4$ has a p-value of 0.3446.

0.3446 = 34.46% of bags contains more than 1200 chocolate chips.

Item d:

This percentile is the p-value of Z when X = 1425, thus:

$Z = (X - \mu)/(\sigma)$

$Z = (1425 - 1252)/(129)$

$Z = 1.34$

$Z = 1.34$ has a p-value of 0.91.

The bag is in the 91st percentile.

A similar problem is given at brainly.com/question/13680644

Final answer:

The probability and percentiles can be found using the z-score formula and then looking these z-scores up in the standard normal distribution table.

Explanation:

In this question, we're dealing with a normal distribution. For a normal distribution, probabilities can be calculated using the z-score formula which is given by Z = (X - μ) / σ, where X is the value from which you want to find the probability, μ is the mean and σ is the standard deviation.

(a) To find the probability that a randomly selected bag has between 1000 and 1500 chocolate chips, we need to find the z-scores for both 1000 and 1500 and then find the area between these two z-scores in the standard normal distribution table. (b) To find the probability that a randomly selected bag has less than 1025 chips, we find the z-score for 1025 and find the area to the left of it in the standard normal distribution table. (c) To find the proportion of bags containing more than 1200 chips, we find the z-score for 1200 and then find the area to the right of it in the standard normal distribution table. (d) Percentile rank can be found by finding the z-score for 1425 chips, and then finding the corresponding percentile in the standard normal distribution table.