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3 years ago

This class is algebra

Mathematics

1 answer:

Contact [7]3 years ago

5 0

Answer:

y = -1/2

Step-by-step explanation:

(3x + 4y = 4)

(2x - 4y = 6)

______________

5x = 10

x = 2

plug in x for any of the two equations ....

2(2) - 4y = 6

4 - 4y = 6

-4y = 2

y = -1/2

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Ostrovityanka [42]

Answer:

i can JUST GET ON YOUR CROME BOOK

Step-by-step explanation:

3 0

3 years ago

There are 63 students in band. This represents 30% of the grade. How many students are in the entire grade

almond37 [142]

So, the question is 63 is 30% of what number?

That can translate to the equation

63 = 30% · x or 63 = 0.3x

Divide by 0.3 and you get x = 210.

There are 210 students in band.

8 0

3 years ago

Read 2 more answers

HELP ASAP 50 POINTS

Nastasia [14]

Answer: x = the quantity of 5 plus or minus the square root of 29 all over 2

Step-by-step explanation:

The given quadratic equation is expressed as

x² - 5x - 1 = 0

The equation is already in the standard form of ax² + bx + c

The general formula for solving quadratic equations is expressed as

x = [- b ± √(b² - 4ac)]/2a

From the given quadratic equation,

a = 1

b = - 5

c = - 1

Therefore,

x = [- - 5 ± √(- 5² - 4 × 1 × - 1)]/2 × 1

x = [5 ± √(25- - 4)]/2

x = [5 ± √29]/ 2

x = ( 5 + √29)/- 2 or x = (5 - √29)/2

4 0

3 years ago

Read 2 more answers

In ΔCDE, the measure of ∠E=90°, the measure of ∠C=70°, and DE = 12 feet. Find the length of CD to the nearest tenth of a foot

Radda [10]

Answer:

$CD = 4.4ft$

Step-by-step explanation:

Given

$DE = 12$

$\angle C = 70^\circ$

Required

Find CD

This question is represented with the attached triangle

To solve CD, we make use of tan formula.

$tan(70) = \frac{DE}{CD}$

Substitute 12 for DE

$tan(70) = \frac{12}{CD}$

Make CD the subject

$CD = \frac{12}{tan(70)}$

$CD = \frac{12}{2.7475}$

$CD = 4.4ft$ --- approximated

6 0

3 years ago

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A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t

julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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

$Z = \frac{X - \mu}{\sigma}$

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with $\mu = \lambda, \sigma = \sqrt{\lambda}$ , if $\lambda>10$ .

Poisson variable with the mean 3

This means that $\lambda= 3$ .

(a) At least 3 in a week.

This is $P(X \geq 3)$ . So

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768$

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

In which

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

$P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680$

$P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008$

$P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504$

$P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216$

Then

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988$