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2 years ago

The table represents a linear function. WhT is the slope of the function?

Mathematics

1 answer:

horrorfan [7]2 years ago

8 0

Answer

The slope of the function is 5.

Step-by-step explanation:

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Four less than the product of ten and a number is eleven. is

nikitadnepr [17]

Answer:

If written in algebraic form it is, 10x - 4 = 11

Step-by-step explanation:

Four less ( minus 4 ) than the product of ten and a number ( 10 times x ) is eleven ( equal to 11 ).

Hopefully this helps!

Brainliest please?

3 0

3 years ago

Power set of {e,f,m,n,o}

Nastasia [14]

$\{\emptyset,\{e\},\{f\},\{m\},\{n\},\{o\},\{e,f\},\{e,m\},\{e,n\},\{e,o\},\{f,m\},\{f,n\},\\
\{f,o\},\{m,n\},\{m,o\},\{n,o\},\{e,f,m\},\{e,f,n\},\{e,f,o\},\{e,m,n\},\\
\{e,m,o\},\{e,n,o\},\{f,m,n\},\{f,m,o\},\{f,n,o\},\{m,n,o\},\{e,f,m,n\},\\
\{e,f,m,o\},\{e,f,n,o\},\{e,m,n,o\},\{f,m,n,o\},\{e,f,m,n,o\}\}$

5 0

3 years ago

Read 2 more answers

What is the value of the remainder if 10x4 – 6x3 + 5x2 – x + 1 is divided by x – 3?

natita [175]

Answer:

Remainder is 691.

Step-by-step explanation:

Given function is $10x^4-6x^3+5x^2-x+1$ .

Now we need to find remainder if we divide given function $10x^4-6x^3+5x^2-x+1$ by (x-3)

(x-3) means plug x=3 into $10x^4-6x^3+5x^2-x+1$ to find remainder.

$10x^4-6x^3+5x^2-x+1$

$=10(3)^4-6(3)^3+5(3)^2-(3)+1$

$=10(81)-6(27)+5(9)-(3)+1$

$=810-162+45-3+1$

$=856-162-3$

$=856-165$

$=691$

Hence remainder is 691.

8 0

3 years ago

Read 2 more answers

To solve the system of equations 6x – 4y = 9 and

mixer [17]

Answer:

4x-2y=6

Step-by-step explanation:

i believe this works

6 0

2 years ago

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

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